Chapter 2: Work, Energy and Power
In the 1840s, a young Englishman named James Prescott Joule spent his evenings doing something that looked pointless. He had grown up around his family’s brewery in Salford, where thermometers mattered, and he had become very good at measuring tiny changes in temperature.
His apparatus was simple. A pair of weights hung on strings, and as they sank they turned a paddle wheel sealed inside a barrel of water. Nothing dramatic happened. The weights fell slowly, the paddles churned, and the water warmed by a fraction of a degree.
But Joule measured that fraction with great care, and then repeated the experiment with different weights, different heights of fall, and even with mercury and oil in place of water. Every time, the answer came out the same. A fixed amount of falling weight always produced a fixed amount of warming.
That was the surprise. The energy of the falling weights had not vanished when the weights stopped. It had turned into heat in the water, in an exact and repeatable ratio. When Joule announced his results in 1845, few people were convinced. Within a few decades, his stubborn measurements had become one of the foundations of physics, and the unit in which we now measure work and energy carries his name.
Diagram
Fig. 2.0-1 — Joule’s Paddle Wheel Experiment Labelled sketch of the apparatus: a pair of weights hanging on strings that pass over pulleys and turn a vertical shaft; the shaft carries a paddle wheel inside a closed, insulated barrel of water; a sensitive thermometer is inserted into the barrel through the lid. Label the falling weights, the height of fall, the pulleys, the paddle wheel, the water, the insulating jacket and the thermometer. Add a small arrow beside the weights marked ‘weights fall’ and a small arrow at the thermometer marked ‘temperature rises slightly’.
2.1 Work
2.1.1 When Is Work Said To Be Done
In everyday speech we use the word ‘work’ very loosely. A student says she has been working all evening while reading a book, and a clerk says he has worked all day while sitting at a desk. Physics uses the word in a much narrower sense.
In Physics, work is said to be done only when a force acting on a body actually makes the body move. The force must produce a displacement. If the body does not move at all, then no matter how hard the force is applied, the work done is zero.
Look at the everyday situations in Fig. 2.1-1 and Fig. 2.1-2. A man pushing a stalled car makes it roll forward, so he does work on it. A cyclist pressing on the pedals, a boy climbing a staircase, a porter raising a trunk onto a rack — in each case a force is applied and something moves, so work is done.
Diagram
Fig. 2.1-1 — A Force That Produces Motion Side view of a person pushing a car from behind on a level road. Show a horizontal arrow at the hands labelled ‘force applied’ and a longer horizontal arrow ahead of the car labelled ‘motion’. Add dashed marks on the road under the wheels to show the car has shifted position. Caption idea for illustrator: the car moves, so work is being done.
Now compare that with a man pushing hard against a thick wall, shown in Fig. 2.1-2. He may strain until his arms ache, but the wall does not shift. Scientifically, he has done no work on the wall. In the same way, a porter standing still with a heavy trunk on his head does no work on the trunk, however tired he feels, because the trunk is not being displaced.
Diagram
Fig. 2.1-2 — A Force That Produces No Motion Side view of a person leaning into a solid brick wall with both hands, feet braced. Show a horizontal arrow at the hands labelled ‘force applied’ and a bold label beside the wall reading ‘no motion’. Show the person’s feet and the wall in the same position, with no displacement marks on the ground.
Feeling tired is not the test. Tiredness comes from the muscles of the body, which keep tightening and relaxing to hold a load in place, and that costs the body energy. But work in the scientific sense is decided by one thing only: did the force move its point of application?
2.1.2 Measurement Of Work
How much work is done? Think about what should make the number bigger. Pushing a trolley harder should count for more, and pushing it further should count for more as well. So the amount of work done depends on two things: the size of the force applied, and the size of the displacement produced in the direction of that force.
If two forces \(F_1\) and \(F_2\), with \(F_1\) greater than \(F_2\), drag two bodies through the same distance, the larger force does the greater work. If the same force \(F\) moves one body through a long distance and another through a short one, more work is done on the first. Both factors matter, and they multiply.
Fig. 2.1-3 shows the simplest case, in which a constant force \(F\) acts on a body and moves it from A to B along the direction of the force itself. The displacement is \(AB = S\).
Diagram
Fig. 2.1-3 — Work Done By A Force Along Its Own Direction A body shown as a small shaded block at position A on a horizontal line, and the same block shown again at position B further to the right. A horizontal arrow at A pointing right is labelled F. A double-headed arrow between A and B along the line is labelled S. Mark the line as the direction of the force.
The work done is then
\[W = F \times S\]
where \(F\) is the constant force applied in newton (N), \(S\) is the displacement of the point of application of the force, measured in the direction of the force, in metre (m), and \(W\) is the work done in joule (J). Two conditions are assumed here: the force does not change while the body moves, and it goes on acting for as long as the body moves.
Read the relation once more with \(S = 0\) in it. If the displacement is zero, then \(W = 0\), whatever the value of \(F\). This is exactly the case of the man and the wall, now written as an equation rather than a story.
2.1.3 Work When The Force Is At An Angle To The Displacement
A force does not always move a body along its own direction. When you pull a suitcase by its handle, your pull slopes upwards while the suitcase slides forward along the floor. The force and the displacement then make an angle \(\theta\) with each other, and only part of the effort goes into moving the body forward.
There are two ways of handling this, and they lead to the same answer. Either we take the part of the displacement that lies along the force, or we take the part of the force that lies along the displacement.
Component Of Displacement Along The Force
In Fig. 2.1-4, a constant force \(F\) acts along AB, while the body is actually displaced along the sloping surface from A to C. The displacement is \(AC = S\), and it makes an angle \(\theta\) with the direction of the force.
Diagram
Fig. 2.1-4 — Taking The Component Of The Displacement Along The Force A right-angled triangle ABC with AB horizontal, B at the right, and C vertically above B. A shaded block sits at A on the sloping line AC and again at C. An arrow at A along AB is labelled F, and a second arrow at C in the same horizontal direction is labelled F. The sloping line AC is labelled ‘displacement S’. Mark the angle θ between AC and AB at A. Draw the perpendicular CB from C to the line AB, and label the segment AB along the base as S cos θ.
Drop a perpendicular CB from C onto the line AB. Then AB is the part of the displacement that lies along the force. From the right-angled triangle ABC,
\[\cos\theta = \frac{\text{base}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{AB}{S}\]
so that \(AB = S\cos\theta\). Since work is force multiplied by the displacement along the force,
\[W = F \times S\cos\theta\]
Component Of Force Along The Displacement
The second route is drawn in Fig. 2.1-5. Here the arrow PA represents the size and direction of the force \(F\) acting on the body, and the body is displaced along AC. Drop a perpendicular from P onto the line AC, meeting it at N. Then NA is the part of the force that lies along the displacement.
Diagram
Fig. 2.1-5 — Taking The Component Of The Force Along The Displacement The same sloping line AC from A up to C, with a shaded block at A and at C and the line labelled ‘displacement S’. From a point P to the lower left of A, draw the arrow PA labelled F, making an angle θ with AC at A. Draw the perpendicular PN from P to the line AC, meeting it at N between P and A, and label the segment NA as F cos θ. Mark the angle θ at A between PA and AC. Also show the force arrow F at C parallel to PA.
From the right-angled triangle PNA, \(NA = PA\cos\theta = F\cos\theta\). The work done is this component multiplied by the whole displacement,
\[W = F\cos\theta \times S\]
which is the same relation as before. In its usual form,
\[W = F S \cos\theta\]
where \(F\) is the constant force in newton (N), \(S\) is the displacement in metre (m), \(\theta\) is the angle between the direction of the force and the direction of the displacement, and \(W\) is the work done in joule (J). The relation holds for a force that stays constant in size and direction while the body moves.
So the work done depends on three things: how big the force is, how big the displacement is, and the cosine of the angle between them. Notice that \(F \times S\) is an ordinary multiplication of two numbers, the magnitudes of the force and the displacement. Work has size but no direction of its own; it is a scalar quantity, even though force and displacement are both vectors.
2.1.4 Special Cases: Positive, Zero And Negative Work
The factor \(\cos\theta\) can be positive, zero or negative, and each of the three possibilities describes a different physical situation. The sign of the work is not bookkeeping; it tells you whether the force is helping the motion, ignoring it, or fighting it.
When The Work Is Positive
If the displacement is along the direction of the force, \(\theta = 0^\circ\) and \(\cos 0^\circ = 1\), so
\[W = F \times S\]
and the work done is positive. A force doing positive work is a force helping the body along.
Fig. 2.1-6 shows the plainest example. A body of mass \(m\) falls freely from A to B through a height \(h\). The force of gravity \(mg\) acts vertically downwards and the displacement \(h\) is also vertically downwards, so the work done by gravity is \(W = FS = mgh\), and it is positive.
Diagram
Fig. 2.1-6 — Positive Work Done By Gravity In Free Fall A small ball at point A near the top and the same ball at point B directly below it. A vertical arrow from the ball at A pointing downwards is labelled F = mg. A second vertical downward arrow beside it is labelled ‘direction of motion (displacement)’. The vertical gap between A and B is marked h with arrowheads at both ends.
A porter raising a load onto a shelf is the same case turned upside down. He must push up with a force equal to \(mg\), and the load moves up through \(h\), so the force he applies and the displacement are both upward and the work he does on the load is \(mgh\), again positive.
When The Work Is Zero
If the displacement is at right angles to the force, \(\theta = 90^\circ\) and \(\cos 90^\circ = 0\), so \(W = 0\). Here the force neither helps nor hinders the motion; it merely acts sideways to it.
A porter walking along level ground with a load on his head is doing no work against gravity, because gravity pulls the load vertically downwards while the load moves horizontally. The two directions are at right angles. He does have to work against friction as he walks, but that is a different force.
A body moving in a circular path in a horizontal plane, shown in Fig. 2.1-8, is the second example. At every instant the force on the body is directed towards the centre of the circle, while the displacement at that instant is along the tangent, at right angles to it. The work done by that force is therefore zero — which is why the speed of a body going round a circle at a steady rate does not change, even though a force is acting on it all the time.
Diagram
Fig. 2.1-8 — Zero Work In Circular Motion A circle with centre O. Mark a point on the circle at the left of O. Draw an arrow from that point towards O along the radius, labelled F. Draw a second arrow from the same point, tangential to the circle and pointing downwards, labelled S. Mark the right angle of 90° between the two arrows at that point.
Putting the two together gives a clear rule: the work done by a force is zero in exactly two situations — when there is no displacement at all (\(S = 0\)), and when the displacement is at right angles to the force (\(\theta = 90^\circ\)).
When The Work Is Negative
If the displacement is opposite to the direction of the force, \(\theta = 180^\circ\) and \(\cos 180^\circ = -1\), so
\[W = -F \times S\]
and the work done on the body by that force is negative. This happens whenever a force opposes the motion or tries to bring a moving body to rest. The minus sign is a statement about the world: this force is taking energy away from the body rather than giving energy to it.
Friction is the everyday case. When a ball rolls across a floor, friction acts backwards along the floor while the ball moves forwards, so the work done on the ball by friction is negative, and the ball slows down.
Gravity does negative work whenever a body rises. In Fig. 2.1-7 a ball of mass \(m\) is thrown upwards from A to B through a height \(h\). The displacement is upwards, the force of gravity \(mg\) is downwards, and the work done on the ball by gravity is \(W = -mgh\).
Diagram
Fig. 2.1-7 — Negative Work Done By Gravity On A Rising Ball A ball at point B near the top and the same ball at point A directly below it on the ground line. A downward arrow at B labelled F = mg and a second downward arrow at A labelled F = mg. Between them a long upward arrow labelled ‘direction of motion (displacement)’. Mark the vertical gap between A and B as h with arrowheads at both ends.
2.1.5 Work Done By A Variable Force
Everything so far has assumed a constant force. In practice a force often changes while the body moves — the pull on a spring grows as it stretches, and the force on a cart changes as the ground changes. For such a case we cannot simply multiply, because there is no single value of \(F\) to multiply by.
The way out is to draw a graph. Plot the force on the Y-axis and the displacement on the X-axis. The area enclosed between the plotted line and the displacement axis then gives the work done.
Fig. 2.1-9 shows a force that increases linearly with displacement, so the graph is a sloping straight line OA through the origin. The area enclosed is the triangle OAB, and its area is \(\tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} F \times S\), which is the work done by the increasing force over the displacement \(S\).
Diagram
Fig. 2.1-9 — Work Done By A Variable Force Graph with the Y-axis labelled ‘force F (N)’ and the X-axis labelled ‘displacement S (m)’. Mark the Y-axis in steps of 5 up to 20 and the X-axis in steps of 2 up to 6. Draw a straight line OA from the origin rising to the point (6, 20). Drop the vertical dashed line from A down to the point B on the X-axis at 6 m, and mark the value 20 on the Y-axis with a dashed horizontal line to A. Shade the triangle OAB and label the shaded area ‘work done’.
This graphical method is not a special trick for variable forces; it works in every case. If the force is constant, the graph is a straight line parallel to the X-axis, as in Fig. 2.1-10, and the area enclosed is the rectangle OABC, whose area is \(F \times S\) — the very relation we started with.
Diagram
Fig. 2.1-10 — Work Done By A Constant Force Graph with the Y-axis labelled ‘force F (N)’ and the X-axis labelled ‘displacement S (m)’. Draw a horizontal straight line AB at a height marked F on the Y-axis, running from A on the Y-axis across to B above the point S on the X-axis. Drop a vertical dashed line from B to the point C on the X-axis, marked S. Shade the rectangle OABC and label the shaded area ‘work done = F × S’.
2.2 Work Done By The Force Of Gravity
Of all the forces you meet, gravity is the one that is always switched on. It pulls every object vertically downwards, and it does so whether the object is falling, sliding, being carried or standing still. It is worth working out, once and for all, how much work gravity does when a body changes its height.
Let a body of mass \(m\) be brought down through a vertical height \(h\). The force of gravity on it is \(F = mg\), acting vertically downwards, where \(m\) is the mass in kilogram (kg) and \(g\) is the acceleration due to gravity in \(\text{m s}^{-2}\). Since the body moves down through \(h\), the displacement in the direction of this force is \(S = h\).
Putting these into \(W = F \times S\) gives
\[W = FS = mgh\]
where \(W\) is the work done by the force of gravity in joule (J), \(m\) is the mass in kilogram (kg), \(g\) is the acceleration due to gravity in \(\text{m s}^{-2}\), and \(h\) is the vertical height through which the body descends, in metre (m). The relation gives the work done by gravity alone; any other force acting at the same time, such as friction on a slope, does its own separate work.
The Path Does Not Matter
Here is the useful part. Suppose the body comes down from the same height \(h\) in three different ways: dropped straight down, slid down a long gentle slope, or carried down a staircase. Fig. 2.2-1 shows all three routes ending at the same level.
Diagram
Fig. 2.2-1 — Three Routes Down The Same Height A single horizontal upper level and a single horizontal lower level, with the vertical gap between them marked h with arrowheads at both ends. From the upper level to the lower level draw three separate paths side by side: on the left a straight vertical dashed arrow labelled ‘dropped straight down’; in the middle a long inclined straight line labelled ‘slid down a slope’; on the right a staircase of five equal steps labelled ‘carried down the stairs’. Place an identical small block at the start of each path and again at the lower level.
In every case the vertical drop is the same, so \(h\) is the same, and the work done by the force of gravity is the same \(mgh\). Only the vertical part of the movement counts, because gravity acts vertically; any sideways part of the journey is at right angles to gravity and contributes nothing, as we saw in section 2.1.4.
For a staircase this is easy to work out. The total height climbed or descended is
\[h = \text{number of steps} \times \text{height of each step}\]
so a flight of 20 steps, each \(15\ \text{cm}\) high, takes you through a vertical height of \(20 \times 0.15\ \text{m} = 3\ \text{m}\). Notice that the step height had to be converted from centimetre to metre first; mixing centimetre with kilogram and second would give a meaningless number.
Going Up Instead Of Down
When a body of mass \(m\) goes up through a vertical height \(h\) — directly, or by stairs, a slope or a lift — the displacement is upward while the force of gravity is downward. The work done by the force of gravity on the body is then
\[W = -mgh\]
and the same amount of work, \(+mgh\), is done by the body against the force of gravity. Both statements describe one event; the minus sign simply records that gravity was opposing the motion rather than assisting it.
2.3 Units Of Work
2.3.1 The S.I. Unit And Its Bigger Units
Work is force multiplied by displacement, so its unit must be a unit of force multiplied by a unit of length. That single sentence fixes the whole system of units for work; nothing else has to be remembered.
In S.I. the unit of force is the newton and the unit of length is the metre, so the S.I. unit of work is the newton metre, which is given the special name joule, abbreviated as J.
Since work \(=\) force \(\times\) displacement,
\[1\ \text{joule} = 1\ \text{newton} \times 1\ \text{metre}\]
Diagram
Fig. 2.3-1 — One Joule Of Work A small block resting on a horizontal surface, drawn twice: once at the starting position and once one metre to the right. A horizontal arrow at the first block pointing right is labelled 1 N. A double-headed arrow along the surface between the two positions is labelled 1 m. Below the sketch place the label ‘work done = 1 J’.
A joule is a small amount of work — roughly what you do in lifting a full glass of water from the table to your mouth. For larger amounts we use multiples of it, made in the usual way with the S.I. prefixes:
- kilo-joule (kJ), where \(1\ \text{kJ} = 10^{3}\ \text{J}\)
- mega-joule (MJ), where \(1\ \text{MJ} = 10^{6}\ \text{J}\)
- giga-joule (GJ), where \(1\ \text{GJ} = 10^{9}\ \text{J}\)
2.3.2 The C.G.S. Unit And Its Relation With The Joule
Before S.I. units became standard, measurements in physics were often made in the C.G.S. system, which uses the centimetre, the gram and the second. Its unit of force is the dyne, so its unit of work is the dyne centimetre, given the special name erg.
\[1\ \text{erg} = 1\ \text{dyne} \times 1\ \text{cm}\]
Relating The Joule And The Erg
The two units measure the same physical quantity, so one can always be converted into the other. Start from the definition of the joule and replace each S.I. unit by its C.G.S. value.
\[1\ \text{J} = 1\ \text{N} \times 1\ \text{m}\]
But \(1\ \text{N} = 10^{5}\ \text{dyne}\) and \(1\ \text{m} = 10^{2}\ \text{cm}\), so
\[1\ \text{J} = 10^{5}\ \text{dyne} \times 10^{2}\ \text{cm} = 10^{7}\ \text{dyne cm}\]
\[1\ \text{J} = 10^{7}\ \text{erg}\]
An erg is therefore a very small unit indeed: it takes ten million of them to make one joule.
2.4 Power
2.4.1 Measurement Of Power
Two pumps are installed to fill the same overhead tank. One fills it in ten minutes, the other takes half an hour. Both have done exactly the same work, because both have raised the same water through the same height. What separates them is how fast they did it. That is what power measures.
If a source does work \(W\) in time \(t\), then
\[P = \frac{W}{t}\]
where \(W\) is the work done in joule (J), \(t\) is the time taken in second (s), and \(P\) is the power in watt (W). This gives the average power over the whole interval; if the rate of working changes from moment to moment, the relation gives the mean value over the time \(t\).
So the power spent by a source depends on two things: how much work it does, and how long it takes to do that work. Doing the same job in less time means spending more power. A machine that finishes in half the time is working at twice the rate.
Fig. 2.4-1 shows the comparison in its simplest form: two porters raising identical loads to the same height of a lorry, one in half the time of the other.
Diagram
Fig. 2.4-1 — Same Work, Different Power Two side-by-side sketches sharing one lorry outline. On the left, a porter lifting a box onto the lorry platform, with a small clock beside him reading 60 s. On the right, a second porter lifting an identical box onto the same platform, with a clock beside him reading 120 s. Mark the same vertical height h with arrowheads on both sketches. Below the pair, place the labels ‘work done: same’ and ‘power spent: twice as much on the left’.
2.4.2 Power In Terms Of Force And Average Speed
There is a second form of the same relation that is often quicker to use. Suppose a constant force \(F\) displaces a body through a distance \(S\) in the direction of the force, in time \(t\). Then the work done is \(W = F \times S\), and the power is
\[P = \frac{W}{t} = \frac{F \times S}{t}\]
But \(S/t\) is the distance covered per second, which is the average speed \(v\). Substituting,
\[P = F \times v\]
where \(F\) is the constant force in newton (N), \(v\) is the average speed of the body in the direction of the force in \(\text{m s}^{-1}\), and \(P\) is the power in watt (W). The relation holds when the force acts along the direction of motion.
This form is useful whenever a vehicle or a machine moves at a steady speed against a steady opposing force. The engine is not making the vehicle go faster; it is spending power simply to keep it going.
2.4.3 Units Of Power
Since power is work divided by time, its unit must be a unit of work divided by a unit of time.
The S.I. unit of power is the watt, abbreviated as W, and
\[1\ \text{watt} = \frac{1\ \text{joule}}{1\ \text{second}} = 1\ \text{J s}^{-1}\]
Household and industrial machines need larger units, and small electronic devices need smaller ones:
- kilowatt (kW), where \(1\ \text{kW} = 10^{3}\ \text{W}\)
- megawatt (MW), where \(1\ \text{MW} = 10^{6}\ \text{W}\)
- gigawatt (GW), where \(1\ \text{GW} = 10^{9}\ \text{W}\)
- milliwatt (mW), where \(1\ \text{mW} = 10^{-3}\ \text{W}\)
- microwatt (\(\mu\)W), where \(1\ \mu\text{W} = 10^{-6}\ \text{W}\)
In the C.G.S. system the unit of power is the erg per second (\(\text{erg s}^{-1}\)). Since \(1\ \text{J} = 10^{7}\ \text{erg}\),
\[1\ \text{W} = 1\ \text{J s}^{-1} = 10^{7}\ \text{erg s}^{-1}\]
Horse Power
One older unit is still in daily use, especially in mechanical engineering and in the ratings of motors and vehicles. It is the horse power, written H.P., and it is related to the watt by
\[1\ \text{H.P.} = 746\ \text{W} = 0.746\ \text{kW}\]
2.4.4 Work Compared With Power
Work and power are related but they answer different questions, and they are among the pair of quantities students most often exchange for one another. Table 2.1 sets them side by side.
2.5 Energy
2.5.1 Energy As Work Done
A body that is able to do work is said to possess energy. Blowing wind can turn the sails of a mill, so wind has energy. Steam pushing a piston can drive a wheel, so steam has energy. In each case we know the energy is there because we can see work being done.
This gives us a way of measuring energy. The energy possessed by a body is measured by the amount of work that the body can do. Energy and work are therefore the same kind of quantity, measured in the same units.
Watch what happens when work is actually done. A body that does work loses some of its capacity to do more work, so its energy decreases. A body on which work is done gains capacity, so its energy increases. Nothing is lost or made in between: whenever work is done, energy is transferred from one body to another, and the amount transferred is exactly the work done. This chain is set out in FC 2.1.
Flowchart
FC 2.1 — Work As A Transfer Of Energy Body A possesses energy → A applies a force on body B and displaces it → work W is done by A on B → energy of A decreases by W → energy of B increases by W
There is one important case in which no energy is transferred at all, even though a force is acting. If the force on a body is at right angles to its displacement, the work done is zero, so no energy passes. A body moving in a circular path is exactly this case: the force on it points towards the centre while the body moves along the tangent, so its energy is not changed by that force.
2.5.2 Units Of Energy
Since energy is measured by the work a body can do, the units of energy are the same as the units of work.
The S.I. unit of energy is the joule (J) and the C.G.S. unit is the erg, with
\[1\ \text{J} = 10^{7}\ \text{erg}\]
Very large and very small amounts of energy have units of their own, chosen to suit the field in which they are used.
Watt Hour And Kilowatt Hour
Since power is the work done per second, multiplying a power by a time gives back an energy. This is the origin of two everyday units of energy.
One watt hour (Wh) is the energy spent by a source of power one watt working for one hour:
\[1\ \text{Wh} = 1\ \text{W} \times 1\ \text{h} = 1\ \text{J s}^{-1} \times 3600\ \text{s} = 3600\ \text{J} = 3.6\ \text{kJ}\]
One kilowatt hour (kWh) is the energy spent by a source of power one kilowatt working for one hour:
\[1\ \text{kWh} = 1000\ \text{J s}^{-1} \times 3600\ \text{s} = 3.6 \times 10^{6}\ \text{J} = 3.6\ \text{MJ}\]
The kilowatt hour is the unit in which household electrical energy is measured and billed, and it is often called simply one ‘unit’ of electricity.
Calorie
Heat energy is usually measured in the calorie. One calorie is the heat energy required to raise the temperature of one gram of water from \(14.5\ ^\circ\text{C}\) to \(15.5\ ^\circ\text{C}\), that is, through \(1\ ^\circ\text{C}\). It is related to the joule by
\[1\ \text{calorie} = 4.18\ \text{J} \qquad \text{or} \qquad 1\ \text{J} = 0.24\ \text{calorie}\]
A bigger unit of heat energy is the kilocalorie, where \(1\ \text{kilocalorie} = 1000\ \text{calorie} = 4180\ \text{J}\).
Electron Volt
At the other extreme, the energy transferred to or from an atomic particle is far too small to be written comfortably in joule, so it is measured in the electron volt (eV).
One electron volt is the energy gained by an electron when it is accelerated through a potential difference of one volt:
\[1\ \text{eV} = \text{charge on an electron} \times 1\ \text{volt} = 1.6 \times 10^{-19}\ \text{C} \times 1\ \text{V}\]
\[1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}\]
2.5.3 Energy Compared With Power
Energy and power are as easily confused as work and power, and for the same reason: one of them counts the total, the other counts the rate. Table 2.2 separates them.
2.6 Mechanical Energy And Its Forms
Energy comes in many kinds, and we shall meet a dozen of them later in this chapter. One kind, however, belongs to the mechanics we have been doing so far, and it is the one to settle first.
The energy possessed by a body because of its state of rest or of motion is called mechanical energy. A stone held above the ground has it because of where it is; the same stone rushing downwards has it because of how fast it is moving.
Mechanical energy therefore exists in two forms:
- potential energy, the energy a body has at rest because of its position, or its size and shape, and
- kinetic energy, the energy a body has because of its motion.
Flowchart
FC 2.2 — The Two Forms Of Mechanical Energy Mechanical energy → branches into two: (i) potential energy → due to position, or to change in size and shape → branches again into gravitational potential energy and elastic potential energy; (ii) kinetic energy → due to motion
The total mechanical energy of a body is simply the sum of the two:
\[\text{total mechanical energy} = U + K\]
where \(U\) is the potential energy of the body in joule (J) and \(K\) is its kinetic energy in joule (J). Both are measured in the same unit, because both are energies.
A body may hold one form, or the other, or both at once. A book lying on a high shelf has potential energy only. A ball rolling along level ground has kinetic energy only. A ball that has been thrown and is still climbing has both, since it is above the ground and moving at the same time. The two sections that follow take each form in turn.
2.7 Potential Energy
A coconut hanging high on a palm tree is doing nothing at all. Yet if it falls it can crack the ground, which means it is able to do work. Its energy is stored, waiting, and it is stored because of where the coconut is.
Any of these bodies has potential energy: a body held at a height above the ground, a wound-up watch spring, a compressed spring, a bent bow, a stretched rubber string. Notice that the list splits neatly in two. In the first case the body has been moved to a new position; in the rest, the body has been squeezed, stretched or bent into a new shape.
In mechanics we therefore meet potential energy of two kinds: gravitational potential energy, which is due to a changed position of the object, and elastic potential energy, which is due to a changed size and shape of the object.
Diagram
Fig. 2.7-1 — Two Kinds Of Potential Energy Two panels side by side. Left panel: a tall palm tree with a coconut still attached near the top; mark the vertical height of the coconut above the ground with a double-headed arrow labelled ‘raised position’; caption label below: gravitational potential energy. Right panel: a rubber band stretched between two fingers, with a small arrow at each end showing the pull, and a dashed outline of the same band unstretched drawn beneath it; caption label below: elastic potential energy.
2.7.1 Gravitational Potential Energy
Every body near the Earth is pulled towards it. That pull is the force due to gravity on the body, which is what we call its weight. Because the Earth attracts it, a body has energy simply by virtue of where it is placed, and this energy is called its gravitational potential energy.
To measure it we need a starting point, a place where we agree the energy is zero. That place is infinity. Very far from the Earth the force of attraction on the body has fallen to zero, so the gravitational potential energy of a body at infinity is zero.
The gravitational potential energy of a body at any point is then defined as the work done by the force of attraction of the Earth in bringing the body from infinity to that point. Here the pulling is done by gravity itself, not by us, and that is the reason for a result which surprises students at first: at any finite distance from the Earth, the gravitational potential energy of a body is negative.
As the body is taken further from the surface of the Earth, the force of attraction on it decreases, and its gravitational potential energy increases — that is, the negative value becomes smaller and smaller, until at infinite separation it finally becomes zero. The negative sign is not a defect in the arithmetic; it is a record of the fact that gravity, not we, did the work of bringing the body in.
2.7.2 Elastic Potential Energy
Some bodies are rigid and hold their shape whatever you do. Others are not: a rubber band, a spring, a strip of steel, a length of bamboo. When an external force is applied to such a non-rigid body, it is deformed, that is, its size and shape change. On removing the force it tends to return to its original form, because of the property of elasticity.
Deforming the body took work, since the applied force moved through a distance while stretching or squeezing it. That work is not lost. It is stored in the body in the form of its elastic potential energy.
So the amount of elastic potential energy stored is exactly the amount of work done in stretching, compressing or bending the body — no more and no less, provided the body is a good elastic one and returns fully to its original form.
2.8 Gravitational Potential Energy At A Height
For everyday problems near the Earth’s surface we do not need infinity as our zero level. It is far more convenient to agree that a body lying on the ground has zero gravitational potential energy, and to measure the energy of a raised body from there.
On that agreement, the gravitational potential energy of a body at a height above the ground is measured by the work done in lifting it up from the ground to that height against the force of gravity. It is denoted by \(U\).
Deriving The Expression
Let a body of mass \(m\) be lifted from the ground to a vertical height \(h\), as in Fig. 2.8-1. To raise it steadily, without giving it any acceleration, the least upward force needed is one that just balances the downward force of gravity on it. That force is \(mg\).
Diagram
Fig. 2.8-1 — Lifting A Body To A Height h A block of mass m shown on the ground line at the bottom and again at a raised position at the top, with a dashed vertical line joining the two positions. At the raised block draw an upward arrow labelled ‘lifting force = mg’ and a downward arrow labelled ‘force of gravity = mg’. Mark the vertical gap between the ground line and the raised block as h with arrowheads at both ends. Label the ground line ‘U = 0 taken here’.
The lifting force \(mg\) acts upwards and the body moves upwards through \(h\), so force and displacement are in the same direction and the work done on the body is
\[W = \text{force of gravity} \times \text{displacement} = mg \times h = mgh\]
This work is not lost when the lifting stops. It has been stored in the body because of its new position, and it is available to be given back if the body falls. So
\[U_h = mgh\]
where \(U_h\) is the gravitational potential energy of the body at height \(h\) in joule (J), \(m\) is its mass in kilogram (kg), \(g\) is the acceleration due to gravity in \(\text{m s}^{-2}\), and \(h\) is the height above the ground in metre (m). Strictly, \(mgh\) is the gain in potential energy when the body is raised through \(h\); we may call it the potential energy at that height only because we have chosen the potential energy on the ground to be zero.
Rising, Falling, And Changing Height
Read the expression as a statement about height and it becomes easy to use. When a body is thrown vertically upwards, its height above the ground increases, so its potential energy increases. When a body is released from a height it falls, its height decreases, so its potential energy decreases, and on our chosen scale it becomes zero at the Earth’s surface.
Most problems ask not for the potential energy at one place but for the change between two. If a body of mass \(m\) is taken up from a height \(h_1\) to a greater height \(h_2\), then
\[\text{gain in potential energy} = mgh_2 - mgh_1 = mg(h_2 - h_1)\]
and if instead it falls from a height \(h_1\) to a lower height \(h_2\), then
\[\text{loss in potential energy} = mgh_1 - mgh_2 = mg(h_1 - h_2)\]
In both cases only the difference in height enters, so the answer does not depend on where the zero level was chosen. Fig. 2.8-2 shows the two levels.
Diagram
Fig. 2.8-2 — Change In Potential Energy Between Two Levels A vertical ground line at the bottom, with two horizontal dashed levels above it. The lower dashed level carries a block labelled ‘mass m at height h₁’ and the upper dashed level carries the same block labelled ‘mass m at height h₂’. Mark h₁ from the ground to the lower level and h₂ from the ground to the upper level, each with arrowheads. Between the two dashed levels mark the gap (h₂ − h₁) and label it ‘height gained’. Add an upward arrow beside the two positions labelled ‘body raised’.
2.9 Kinetic Energy
A stone lying on the road is harmless. The same stone thrown hard can break a window pane. Nothing about the stone has changed except that it is now moving, so the energy must have come from the motion itself.
Every moving thing has kinetic energy: a bullet leaving a gun, the hands of a clock creeping round the dial, an apple falling from a branch, a train pulling out of a platform. In each case the body is able to do work on whatever it meets, and it can go on doing work until it comes to rest.
2.9.1 Expression For Kinetic Energy
That last sentence is the key to measuring kinetic energy. The kinetic energy possessed by a moving body is equal to the amount of work the body is capable of doing before coming to rest. So instead of measuring the energy directly, we work out how much work is needed to stop the body.
Suppose a body of mass \(m\) is moving with a velocity \(v\), as in Fig. 2.9-1. It is brought to rest by a constant opposing force \(F\), which produces a uniform retardation \(a\), and the body travels a distance \(S\) before stopping.
Diagram
Fig. 2.9-1 — Stopping A Moving Body A horizontal surface with a block drawn at the left labelled ‘mass m’ and a velocity arrow pointing right labelled v. The same block drawn again at the right, at rest, labelled ‘at rest’. A horizontal arrow pointing left, against the motion, is drawn at the block and labelled ‘opposing force F’. A double-headed arrow along the surface between the two positions is labelled S. Add a small note beside the arrows: force opposes motion.
The retarding force is
\[F = \text{mass} \times \text{retardation} = ma \qquad \text{...(i)}\]
and the kinetic energy of the body is the work done by this force over the distance \(S\),
\[K = F \times S \qquad \text{...(ii)}\]
To find \(S\), use the equation of motion for uniform acceleration. Here the initial velocity is \(v\), the final velocity is zero, and since \(a\) is a retardation the acceleration in the equation is \(-a\). From \(v_{\text{final}}^{2} = v_{\text{initial}}^{2} + 2 \times \text{acceleration} \times S\),
\[0 = v^{2} - 2aS \qquad \text{so} \qquad S = \frac{v^{2}}{2a} \qquad \text{...(iii)}\]
Substituting (i) and (iii) into (ii),
\[K = ma \times \frac{v^{2}}{2a} = \tfrac{1}{2}mv^{2}\]
\[K = \tfrac{1}{2}mv^{2}\]
where \(K\) is the kinetic energy in joule (J), \(m\) is the mass of the body in kilogram (kg), and \(v\) is its velocity in \(\text{m s}^{-1}\). The derivation assumed a constant opposing force and therefore a uniform retardation, but the result itself is general.
In words, the kinetic energy of a body is half the product of its mass and the square of its velocity. Because the velocity is squared, kinetic energy grows very fast with speed. Double the velocity and the kinetic energy becomes four times as great. Of two bodies of the same mass, the faster one always carries the more energy.
2.9.2 Kinetic Energy And Momentum
Both kinetic energy and momentum describe a moving body, and they are connected. Let a body of mass \(m\) move with velocity \(v\). Then
\[K = \tfrac{1}{2}mv^{2} \qquad \text{...(i)}\]
\[p = mv \qquad \text{...(ii)}\]
where \(p\) is the momentum in \(\text{kg m s}^{-1}\). From (ii), \(v = p/m\). Substituting this into (i),
\[K = \tfrac{1}{2}m\left(\frac{p}{m}\right)^{2} = \frac{p^{2}}{2m}\]
so that \(p^{2} = 2mK\), and the two relations may be written
\[p = \sqrt{2mK} \qquad \text{or} \qquad K = \frac{p^{2}}{2m}\]
These let us compare two bodies of different masses. Take a light body A of mass \(m\) and a heavy body B of mass \(M\), with \(M\) greater than \(m\).
If both have the same momentum \(p\), then
\[\frac{K_A}{K_B} = \frac{p^{2}/2m}{p^{2}/2M} = \frac{M}{m}\]
Since \(M\) is greater than \(m\), this ratio is greater than one, so \(K_A\) is greater than \(K_B\): of two bodies with the same momentum, the lighter one has the greater kinetic energy.
If instead both have the same kinetic energy \(K\), then
\[\frac{p_B}{p_A} = \frac{\sqrt{2MK}}{\sqrt{2mK}} = \sqrt{\frac{M}{m}}\]
which is again greater than one, so \(p_B\) is greater than \(p_A\): of two bodies with the same kinetic energy, the heavier one has the greater momentum.
2.9.3 The Work–Energy Theorem
When a force is applied in the direction in which a body is already moving, the body accelerates, so its kinetic energy increases. The theorem below says exactly how much it increases by.
Proof Of The Theorem
Let a body of mass \(m\) be moving with an initial velocity \(u\). A constant force \(F\) is applied along its direction of motion, producing an acceleration \(a\), and the velocity increases from \(u\) to \(v\) while the body moves a distance \(S\).
\[F = ma \qquad \text{...(i)}\]
\[W = F \times S \qquad \text{...(ii)}\]
From the equation of motion for uniform acceleration, \(v^{2} = u^{2} + 2aS\), so
\[S = \frac{v^{2} - u^{2}}{2a} \qquad \text{...(iii)}\]
Substituting (i) and (iii) into (ii),
\[W = ma \times \frac{v^{2} - u^{2}}{2a} = \tfrac{1}{2}m(v^{2} - u^{2})\]
\[W = \tfrac{1}{2}mv^{2} - \tfrac{1}{2}mu^{2}\]
Now \(\tfrac{1}{2}mu^{2}\) is the initial kinetic energy \(K_i\) of the body and \(\tfrac{1}{2}mv^{2}\) is its final kinetic energy \(K_f\), so
\[W = K_f - K_i\]
that is, the work done on the body equals the increase in its kinetic energy. Here \(W\) is in joule (J), \(m\) in kilogram (kg), and \(u\) and \(v\) in \(\text{m s}^{-1}\); the proof assumes a constant force acting along the direction of motion.
The theorem also works in reverse. If the force opposes the motion, as friction does, the work done on the body is negative and its kinetic energy decreases by that amount.
2.9.4 Forms Of Kinetic Energy
A body can move in three different ways, so kinetic energy comes in three forms.
Diagram
Fig. 2.9-2 — The Three Forms Of Kinetic Energy Three equal panels in a row, each with a caption line beneath it. Left panel: an arrow-straight track with a small trolley on it and a horizontal arrow ahead of it; caption ‘translational — motion in a straight line’. Middle panel: a spinning top upright on a floor with a curved arrow around its axis; caption ‘rotational — motion about an axis’. Right panel: a steel strip clamped at its lower end in a vice, with the free upper end drawn in two dashed positions to left and right of the centre and a double-headed curved arrow between them; caption ‘vibrational — to and fro about a mean position’.
Translational Kinetic Energy
The motion of a body in a straight line path is called translational motion, and the kinetic energy of a body due to such motion is called translational kinetic energy. A bus running along a straight road, a freely falling body, and a molecule of a monoatomic gas all have translational kinetic energy.
Rotational Kinetic Energy
If a body rotates about an axis, the motion is called rotational motion, and the kinetic energy due to it is called rotational kinetic energy, or simply rotational energy. A spinning top, a turning grindstone and the blades of a fan all have it. The Earth, rotating on its own axis, has an enormous amount of it. The atoms of a moving diatomic or polyatomic molecule rotate about a fixed axis as well as travelling along, so such a molecule has both rotational and translational kinetic energy.
Vibrational Kinetic Energy
If a body moves to and fro about a mean position, the motion is called vibrational motion, and the kinetic energy due to it is called vibrational kinetic energy, or simply vibrational energy. A wire clamped at both ends and struck in the middle vibrates, and the energy it then possesses is vibrational kinetic energy. A steel strip clamped at one end vibrates in the same way when its free end is displaced slightly and released. In a solid, the atoms vibrate about their mean positions, so they too possess vibrational kinetic energy, and the molecules of a polyatomic gas can vibrate about their mean positions in addition to rotating and travelling.
Depending on how it is moving, a body may possess more than one of these forms at the same time. The wheel of a moving vehicle is travelling forward and turning at once, so it has translational and rotational kinetic energy together.
2.9.5 Potential Energy Compared With Kinetic Energy
The two forms of mechanical energy are easily told apart once the difference is set out plainly. Table 2.3 does that.
2.10 Conversion Of Potential Energy Into Kinetic Energy
Potential energy is stored energy, and stored energy is of no use until it is released. The moment it is put to use, it changes into kinetic energy.
Four everyday cases show the change happening, two of them gravitational and two elastic.
A Weight Raised And Dropped
On a building site, a pile driver raises a heavy iron block and lets it fall onto the head of a post. Raising the block through a height against the force of gravity gives it gravitational potential energy equal to the work done in lifting it. As soon as the block is released it begins to fall, its height decreases, and its potential energy begins to change into kinetic energy. Part-way down it has some of each. On striking the post the moving block does work on the post, and the post is driven into the ground.
A Wound-Up Watch Spring
A wound-up watch spring has elastic potential energy equal to the work done in bringing the spring to the wound-up state. As the spring unwinds itself, the stored potential energy changes into kinetic energy, which does work in moving the hands of the watch, and so becomes the kinetic energy of the hands.
A Drawn Bow
When the string of a bow is pulled back, work is done on the bow, and it is stored in the deformed bow as elastic potential energy. On releasing the string, this potential energy changes into the kinetic energy of the arrow, which makes the arrow fly. The chain of changes is set out in FC 2.2.
Flowchart
FC 2.2 — Energy Changes In Shooting An Arrow Work done by the archer in pulling the string → elastic potential energy stored in the bent bow → kinetic energy of the arrow leaving the bow → work done by the arrow on the target as it stops
A Compressed Spring
A compressed spring has elastic potential energy because of its compressed state. If a small ball is placed on top of it, as in Fig. 2.10-1, and the spring is then released, the potential energy of the spring changes into kinetic energy, which does work on the ball. The ball gains kinetic energy and flies away.
Diagram
Fig. 2.10-1 — A Ball Launched By A Compressed Spring Two panels side by side, both showing a coiled spring standing upright on a fixed hatched base. Left panel: the spring is squeezed short, held down by a hand, with a small ball resting on its top; label ‘compressed spring, elastic potential energy stored’. Right panel: the same spring extended to its full length with no hand, and the ball drawn above it with an upward arrow; label ‘released spring, ball flies away with kinetic energy’.
In all four cases the pattern is the same. Work was done first to store the energy — lifting, winding, pulling, squeezing — and that stored energy is later handed back as motion. Nothing is created along the way; the energy is only shifted from one form into the other.
2.11 Different Forms Of Energy
Nature supplies us with energy in many forms, and each has its own store, its own carrier and its own uses. Twelve of them are described below. Mechanical energy, the twelfth, has already been treated in section 2.6.
Solar Energy
The energy radiated out by the Sun is called solar energy. It cannot be used directly to perform work, because it arrives too spread out and is not always available uniformly. A number of devices have therefore been invented to make use of it.
- A solar panel consists of a black-painted metal surface which absorbs the Sun’s energy and heats water in contact with it.
- A solar furnace has a large parabolic mirror of wide aperture which focuses the Sun’s rays onto a small area, so that a high temperature is reached there. Water stored at the focus of the mirror can be boiled to produce steam, and the steam can then drive the turbine of an electric generator to produce electrical energy.
- Solar cells convert solar energy directly into electrical energy. They are made from thin slices of a semiconducting material. The efficiency of a solar cell in converting solar energy into electrical energy is, however, very low.
Diagram
Fig. 2.11-1 — A Solar Furnace A large concave parabolic mirror shown in section, facing upwards and to the left. Draw four parallel rays coming in from the upper left labelled ‘rays from the Sun’, striking the mirror surface and reflecting so that all four meet at a single point in front of the mirror. Mark that point ‘focus’ and place a small closed vessel of water there, with a pipe leading away from it labelled ‘steam to turbine’. Label the mirror ‘parabolic mirror of large aperture’.
Heat Energy
The energy released on burning coal, oil, wood or gas is called heat energy. Steam obtained by heating water possesses heat energy and has the capacity to do work; in a steam engine the heat energy of steam is used to obtain mechanical energy. The Sun is also a source of heat energy, and the Earth receives heat energy directly from it.
Light Energy
Light energy is the form of energy in the presence of which other objects are seen. The natural source of light energy is the Sun. The Moon reflects the sunlight falling on it and is thus seen at night. Many sources such as a fire, a burning candle and the heated filament of a bulb give out light energy along with heat energy.
Chemical Energy
The energy possessed by fossil fuels such as coal, petroleum and natural gas is called chemical energy, or fuel energy. These fuels contain chemical energy stored in them, and when they are burnt, the chemical energy changes into heat and light energy.
The food we eat also has chemical energy in it. Plants convert solar energy into the chemical energy of food during the process of photosynthesis. A battery has chemical energy stored in it, and it provides electrical energy when in use, by a chemical reaction which takes place inside it.
Hydro Energy
The energy possessed by fast moving water is called hydro energy. It is used to generate electricity in hydroelectric power stations. Water is stored in dams built across rivers high up in the hills, and water from the dams is allowed to flow down through channels; the kinetic energy of the fast flowing water then drives the turbines of the generators.
Electrical Energy
When two dry bodies are rubbed together they get charged, because free electrons move from one body to the other, so they possess electrical energy. In an electric cell, the movement of ions from one electrode to the other within the cell, as a result of a chemical reaction, becomes a source of electrical energy.
Nuclear Energy
The energy released due to loss in mass during the processes of nuclear fission and fusion is called nuclear energy, or atomic energy. In both these processes the loss in mass gets converted into energy in accordance with Einstein’s mass–energy relation
\[E = mc^{2}\]
where \(E\) is the energy released in joule (J), \(m\) is the mass lost in kilogram (kg), and \(c\) is the speed of light in vacuum, \(3 \times 10^{8}\ \text{m s}^{-1}\). Because \(c^{2}\) is such an enormous number, a very small loss of mass yields a very large amount of energy.
This energy is used in nuclear reactors for constructive purposes, to produce electrical energy. The same energy released in an uncontrolled way is destructive, which is why nuclear technology is closely regulated. Countries differ on how large a part nuclear power should play in their electricity supply: it produces no smoke or carbon dioxide while running, but the disposal of radioactive waste and the cost and safety of the plants are matters of continuing public debate.
Geo-Thermal Energy
The energy released in nuclear disintegrations in the interior of the Earth gets stored deep inside the Earth and is called geo-thermal energy. This energy heats up the underground water to produce natural steam. Sometimes this natural steam bursts out from the surface of the Earth as hot springs. Electric power companies may drill wells into the Earth to trap the natural steam, which can then be used to run the turbines of generators to produce electricity.
Wind Energy
The energy possessed by fast moving air is called wind energy. It is used to drive a wind mill. In rural areas wind mills have for many years been used to pump out underground water and to grind grain. Nowadays, giant wind mills are used to drive the turbines of electric generators for producing electricity.
Sound Energy
A vibrating body possesses mechanical energy capable of producing sound energy, which is sensed by our ears. When the disturbance produced by a vibrating body travels through the layers of the atmosphere and reaches our ears, it produces vibrations in the ear membrane, and sound is heard.
Magnetic Energy
The energy possessed by a magnet, due to which it can attract iron filings, is called magnetic energy. An electromagnet has magnetic energy.
2.12 Conversion Of One Form Of Energy Into Another
In daily life we need energy in a particular form at a particular place — light in a room, motion in a fan, heat in a kettle. We rarely have it in that form to begin with. What makes modern life possible is that one form of energy can be converted into another, so we can obtain energy in the form we want from whatever form we happen to have.
Fig. 2.12-1 sets out some of the commonest conversions and the device that brings each one about. Notice how many of them work in both directions, with a different device for each direction.
Diagram
Fig. 2.12-1 — Some Common Energy Conversions And Their Devices A chart of paired boxes. Each pair has a left box and a right box joined by two horizontal arrows, one pointing right with the name of the device written above it and one pointing left with the name of the other device written below it. Draw six pairs, one under the other. Pair 1: mechanical energy ⇄ electrical energy; right arrow labelled ‘dynamo (generator)’, left arrow labelled ‘motor’. Pair 2: electrical energy → heat energy; arrow labelled ‘electric iron’; return arrow labelled ‘thermocouple’. Pair 3: electrical energy ⇄ sound energy; right arrow labelled ‘loudspeaker’, left arrow labelled ‘microphone’. Pair 4: electrical energy ⇄ chemical energy; right arrow labelled ‘charging a battery’, left arrow labelled ‘cell in use’. Pair 5: chemical energy ⇄ light energy; right arrow labelled ‘lighted candle’, left arrow labelled ‘photosynthesis’. Pair 6: electrical energy ⇄ light energy; right arrow labelled ‘lighted electric bulb’, left arrow labelled ‘photocell’.
2.12.1 Common Energy Conversions Around Us
From Mechanical Energy, And Back To It
Water stored in the reservoir of a dam has potential energy. When the water flows down, its potential energy decreases and its kinetic energy increases. If the falling water is made to rotate a turbine near the bottom of the dam, the kinetic energy of the water is transferred to the turbine as rotational kinetic energy. The turbine rotates the armature of the generator connected to it, and the kinetic energy is transformed into electrical energy in the generator. Thus an electric generator, or dynamo, converts mechanical energy into electrical energy. FC 2.3 follows the chain end to end.
Flowchart
FC 2.3 — Energy Changes In A Hydroelectric Power Station Potential energy of water stored high in the reservoir → kinetic energy of the water falling through the channel → rotational kinetic energy of the turbine → mechanical energy of the generator armature → electrical energy in the transmission line
The reverse change is made by an electric motor. When an electric current is passed through a coil freely suspended or pivoted in a magnetic field, a turning effect acts on the coil and it rotates, carrying the shaft attached to it round with it. So the electrical energy supplied to an electric motor changes into mechanical energy. Electric motors run fans, washing machines, mixers and grinders in the home, and countless machines in industry.
Mechanical energy also changes into heat energy. When water falls from a height, its potential energy changes into kinetic energy during the fall, and on striking the ground a part of that kinetic energy changes into heat energy, so the temperature of the water rises. The moving parts of a machine get heated for the same reason, as friction turns part of their mechanical energy into heat.
Conversions In Electrical Devices
Electrical energy changes into heat energy in appliances such as a heater, an oven, a geyser or a toaster, where a current passes through a resistance wire or a filament. The reverse change takes place in a thermocouple: when its two junctions of two different metals are kept at different temperatures, one hot and one cold, a current flows in it, so the heat energy supplied at the hot junction has been changed into electrical energy.
Electrical energy changes into sound energy in a loudspeaker, which receives electrical signals and gives out sound, and in an electric bell. The reverse change takes place in a microphone, which converts sound energy into electrical energy in the form of varying electric signals.
Electrical energy changes into chemical energy while a battery is being charged, and the stored chemical energy changes back into electrical energy when current is drawn from the cell.
Electrical energy changes into light energy in a glowing electric bulb, though a large part of the supplied energy leaves as heat as well. The reverse change takes place in a photoelectric cell and in a solar cell, in which light energy is converted into electrical energy.
Electrical energy changes into magnetic energy while an electromagnet is being made: a current is passed through a coil wound around a soft iron bar, and the bar gets magnetised.
Conversions Involving Chemical Energy
Chemical energy changes into light energy when a candle burns, and in the same way when the oil soaked in the wick of a kerosene lamp burns.
Light energy changes into chemical energy in photosynthesis: light energy from the Sun is absorbed by green plants, which change it into the chemical energy of food.
Chemical energy changes into heat energy when a fuel such as wood, coal or bio-gas burns. In a steam engine the chemical energy of coal on burning changes into the heat energy of steam. In the explosion of crackers, chemical energy changes into heat, light and sound energies together. In lighting a candle or a match stick, chemical energy changes into heat and light energies, and in respiration inside the body, chemical energy is converted into heat energy.
Chemical energy changes into mechanical energy in an automobile in motion, where the chemical energy of petrol or diesel becomes the kinetic energy of the vehicle.
Conversions Involving Heat And Nuclear Energy
Heat energy changes into mechanical energy in a steam engine, where the heat energy of the steam changes into the kinetic energy of the piston.
Nuclear energy changes into electrical energy in a nuclear reactor, where the energy released in the process of nuclear fission is converted into electrical energy.
One Rule Worth Remembering
Whenever mechanical energy changes into some other form, it always does so as kinetic energy and never directly as potential energy. Stored potential energy first changes into kinetic energy, and the kinetic energy then changes into the desired form. The dam is the clearest illustration: the water’s potential energy must first become the kinetic energy of falling water before the turbine can make anything of it.
2.12.2 Degradation Of Energy
No conversion is perfect. While transforming energy from one form into another desired form, the whole of it never changes into the form we want. A part changes into some other undesirable form, usually heat due to friction, or a part is lost to the surroundings by radiation, and is thus of no use to us.
Two familiar examples show how large the degraded part can be. When a bulb is lighted using electrical energy, a major part of the energy is lost in the form of heat energy, and only a small part is converted into the useful light energy we wanted. When a vehicle is run using the chemical energy of its fuel, a major part is wasted as heat and sound, and only a part changes into useful mechanical energy.
Notice carefully what degradation does not mean. The energy has not been destroyed; every joule is still there. It has simply spread out into forms and places from which we cannot conveniently collect it again.
2.13 Principle Of Conservation Of Energy
Section 2.12 showed energy shifting from form to form, again and again, in devices of every kind. Through all those changes one quantity never altered: the total. That fact is one of the fundamental principles of nature.
In daily life energy occurs in many forms at once, and the sum of all the forms remains constant. When there is a transformation of energy from one form into another, the total energy always remains the same, that is, it remains conserved. If a bulb gives out less light energy than the electrical energy supplied to it, the missing joules have not disappeared; they have left as heat, and if they are counted the books balance exactly.
Diagram
Fig. 2.13-1 — The Energy Account Of A Glowing Bulb A horizontal bar on the left labelled ‘electrical energy supplied to the bulb’. From its right-hand end draw two arrows into two separate bars stacked one above the other: a short upper bar labelled ‘useful light energy’ and a long lower bar labelled ‘heat energy given to the surroundings’. Show with a brace that the two output bars together are exactly as long as the input bar, and label the brace ‘total energy unchanged’.
Conservation Of Mechanical Energy
There is an important special case. Suppose the only changes taking place are between the two forms of mechanical energy, so that potential energy is being traded for kinetic energy and nothing else is happening. Then the total mechanical energy stays constant:
\[K + U = \text{constant}\]
where \(K\) is the kinetic energy in joule (J) and \(U\) is the potential energy in joule (J). This holds only in the absence of all kinds of frictional forces. If friction or air resistance is at work, some of the mechanical energy is converted into heat and sound, and the sum \(K + U\) falls, even though the total energy of every kind is still conserved.
2.14 Conservation Of Mechanical Energy For A Freely Falling Body
The statement \(K + U = \text{constant}\) deserves to be checked rather than believed. The simplest case to check it on is a body falling freely under gravity, because there the height and the speed can both be worked out at any stage.
Let a body of mass \(m\) fall freely under gravity from a height \(h\) above the ground, that is, from position A in Fig. 2.14-1. As the body falls, its potential energy changes into kinetic energy. We shall calculate the sum of the kinetic energy \(K\) and the potential energy \(U\) at three places: at A, at a point B after it has fallen through a distance \(x\), and at C on the ground.
Diagram
Fig. 2.14-1 — Free Fall Of A Body A vertical line with three marked positions. At the top, position A with a small ball, labelled u = 0 and marked ‘K = 0, U = mgh’. Below it, position B with the same ball, marked ‘K = mgx, U = mg(h − x)’ and carrying the velocity label v₁. At the bottom, on the ground line, position C with the ball, marked ‘K = mgh, U = 0’ and carrying the velocity label v. Mark the distance from A to B as x, the distance from B to the ground as h − x, and the whole distance from A to the ground as h, each with arrowheads. Add a downward arrow alongside labelled ‘direction of motion’.
At Position A, At A Height h Above The Ground
The body starts from rest, so its initial velocity is zero and
\[K = 0, \qquad U = mgh\]
\[\text{total energy} = K + U = 0 + mgh = mgh \qquad \text{...(i)}\]
At Position B, After Falling Through A Distance x
Let \(v_1\) be the velocity acquired by the body at B. Here the initial velocity \(u = 0\), the distance fallen \(S = x\), and the acceleration \(a = g\). From \(v^{2} = u^{2} + 2aS\),
\[v_1^{2} = 0 + 2gx = 2gx\]
so the kinetic energy at B is
\[K = \tfrac{1}{2}mv_1^{2} = \tfrac{1}{2}m \times 2gx = mgx\]
The body is now at a height \(h - x\) above the ground, so its potential energy is \(U = mg(h - x)\), and
\[\text{total energy} = mgx + mg(h - x) = mgh \qquad \text{...(ii)}\]
At Position C, On The Ground
Let \(v\) be the velocity acquired on reaching the ground. Here \(u = 0\), \(S = h\) and \(a = g\), so
\[v^{2} = 0 + 2gh = 2gh\]
\[K = \tfrac{1}{2}mv^{2} = \tfrac{1}{2}m \times 2gh = mgh\]
and the potential energy on the ground is zero, since that is the level we chose as zero. Hence
\[\text{total energy} = mgh + 0 = mgh \qquad \text{...(iii)}\]
What The Three Results Show
Equations (i), (ii) and (iii) give the same answer, \(mgh\). So the total mechanical energy — the sum of the kinetic energy and the potential energy — remains constant at every point of the motion, and is equal to the initial potential energy at the height \(h\).
As the body falls, its potential energy decreases and its kinetic energy increases by exactly the same amount. Just at the instant when it strikes the ground, the whole of the potential energy has changed into kinetic energy, so the kinetic energy of the body on reaching the ground equals the potential energy it had at the height \(h\).
A Body Thrown Vertically Upwards
The same account runs backwards for a body thrown vertically upwards. Its initial kinetic energy, supplied at the instant of throwing, keeps on decreasing, and its potential energy keeps on increasing by the same amount. When the body reaches the highest point, the whole of its initial kinetic energy has changed into potential energy, so the body momentarily comes to rest. It is still under the force of gravity, so it at once starts falling, and its potential energy begins to change back into kinetic energy.
This gives a useful result. The initial kinetic energy provided to a body at the ground, so that it may reach a certain height \(h\), must equal the potential energy it will have at that height. So
\[\tfrac{1}{2}mu^{2} = mgh \qquad \text{or} \qquad u = \sqrt{2gh}\]
where \(u\) is the velocity of projection in \(\text{m s}^{-1}\), \(g\) is the acceleration due to gravity in \(\text{m s}^{-2}\), and \(h\) is the greatest height reached in metre (m). The relation assumes that no energy is lost against air friction.
The Same Account As A Graph And As A Table
Fig. 2.14-2 shows how the kinetic energy and the potential energy of the body vary with its height above the ground.
Diagram
Fig. 2.14-2 — Kinetic And Potential Energy Against Height Graph with the Y-axis labelled ‘energy (J)’ and the X-axis labelled ‘height above the ground (m)’. Mark mgh and ½mgh on the Y-axis, and mark the points C at 0, B at h/2 and A at h along the X-axis. Draw a horizontal line right across the top at the level mgh, labelled ‘total energy’. Draw a straight line starting at the value mgh on the Y-axis at height 0 and falling to zero at height h, labelled ‘kinetic energy’. Draw a second straight line starting at zero at height 0 and rising to mgh at height h, labelled ‘potential energy’. Mark the crossing point of the two sloping lines with a dashed vertical line down to h/2 on the X-axis.
Table 2.4 sets out the same information in numbers, for a body of mass \(m\) moving vertically downwards and vertically upwards.
Table 2.4 — Kinetic energy and potential energy of a body in vertical motion
| Motion | Height above the ground | Kinetic energy \(K\) | Potential energy \(U\) | Total energy \(E = K + U\) |
|---|---|---|---|---|
| Downward motion (free fall) | \(h\) (highest point A) | \(0\) | \(mgh\) | \(mgh\) |
| Downward motion (free fall) | \(\tfrac{1}{2}h\) (middle point B) | \(\tfrac{1}{2}mgh\) | \(\tfrac{1}{2}mgh\) | \(mgh\) |
| Downward motion (free fall) | \(0\) (ground C) | \(mgh\) | \(0\) | \(mgh\) |
| Upward motion | \(0\) (ground C) | \(mgh\) | \(0\) | \(mgh\) |
| Upward motion | \(\tfrac{1}{2}h\) (middle point B) | \(\tfrac{1}{2}mgh\) | \(\tfrac{1}{2}mgh\) | \(mgh\) |
| Upward motion | \(h\) (highest point A) | \(0\) | \(mgh\) | \(mgh\) |
Where This Account Is Not Quite True
In all these calculations the force of friction between the body and the air has been ignored. In fact, during the fall some of the kinetic energy changes into heat energy because of air friction, and that heat is dissipated into the air. So at the ground C the kinetic energy will be a little less than \(mgh\).
The conservation of mechanical energy is therefore strictly valid only in the absence of external forces such as air friction — although the total sum of energy of all kinds, mechanical and otherwise, is always the same, that is, conserved.
2.15 Conservation Of Energy In A Simple Pendulum
A simple pendulum is nothing more than a small heavy bob hung by a light thread from a rigid support. Yet it is the neatest demonstration of conservation of mechanical energy available, because the same trade between potential energy and kinetic energy is repeated over and over in front of your eyes.
Fig. 2.15-1 shows such a pendulum, hung from a rigid support O, with its resting position at A. When the bob is displaced to one side and then released, it swings from one side to the other, reaching an equal distance and an equal height on either side. Neglecting the force of friction between the bob and the surrounding air, the motion can be explained entirely by the principle of conservation of energy.
Diagram
Fig. 2.15-1 — Energy Changes In A Simple Pendulum A rigid horizontal support hatched along the top, with the point of suspension O marked on it. From O draw three straight thread lines: one vertical down to the bob at the mean position A, one sloping to the left down to the bob at the extreme position C, and one sloping to the right down to the bob at the extreme position B. Draw a shallow arc through C, A and B to show the path of the bob. Mark the vertical height of B above A as h, and the vertical height of C above A as h, each with a small double-headed arrow and a dashed horizontal line through A. Label position B ‘P.E. = mgh, K.E. = 0’, label position C ‘P.E. = mgh, K.E. = 0’, and label position A ‘P.E. = 0, K.E. = mgh’.
Following The Bob Through One Swing
Let A be the resting, or mean, position of the bob, where we take its potential energy to be zero. At the mean position the bob is at its lowest, so its potential energy is the least it can be, and since only the change in potential energy matters we are free to call that least value zero.
When the bob is displaced from A to B, it gets raised through a vertical height \(h\), so its potential energy increases by \(mgh\), where \(m\) is the mass of the bob in kilogram (kg). At B it is momentarily at rest, so all its energy is potential energy.
On releasing the bob at B, it moves back from B to A. Its vertical height decreases from \(h\) to zero, so its potential energy decreases from \(mgh\) to zero, and that energy is converted into kinetic energy:
\[\tfrac{1}{2}mv^{2} = mgh \qquad \text{so} \qquad v = \sqrt{2gh}\]
At A, therefore, the bob has no potential energy left and is moving fastest, with velocity \(v = \sqrt{2gh}\) in \(\text{m s}^{-1}\). Being in motion, it carries on past A towards C.
As the bob rises from A to C its kinetic energy is spent again in gaining height. At C, having risen through the same vertical height \(h\) above A, it once more has potential energy \(mgh\) and its kinetic energy has become zero, so the bob momentarily comes to rest at C. But the force of gravity is still acting on it, so it at once begins to move back from C towards A.
As the bob swings back from C to A, its potential energy decreases and its kinetic energy increases. At A it again has the whole of its mechanical energy in the form of kinetic energy, with zero potential energy, so it swings on from A to B and the whole process repeats.
Flowchart
FC 2.4 — Energy Changes During One Complete Swing Bob at B, all energy potential → bob at A, all energy kinetic → bob at C, all energy potential → bob at A, all energy kinetic → back to B, where the cycle begins again
The Pattern In One Sentence
During the swing, at the extreme positions B and C the bob has only potential energy, while at the mean position A it has only kinetic energy. At any intermediate position, between A and B or between A and C, the bob has both kinetic energy and potential energy, and the sum of the two — the total mechanical energy — remains constant throughout the swing.
This last statement is strictly true only in vacuum, where there is no force of friction due to air. In a real pendulum a little of the mechanical energy is converted into heat and sound at every swing, so the bob rises to a slightly lower height each time and the pendulum finally comes to rest. Even then, no energy has been destroyed; it has simply passed into the surrounding air in a degraded form.
2.16 Summary
Work is said to be done by a force only when the force applied on a body makes the body move, that is, when there is a displacement. Work is a scalar quantity.
The work done by a force is the product of the force and the displacement of the point of application of the force in the direction of the force. For a constant force acting along the displacement, \[W = F \times S\] where \(F\) is the force in newton (N), \(S\) is the displacement in metre (m) and \(W\) is the work done in joule (J).
When the force is at an angle to the displacement, \[W = FS\cos\theta\] where \(\theta\) is the angle between the directions of the force and the displacement. The relation holds for a constant force.
Work done is positive when \(\theta = 0^\circ\), so that \(W = FS\); the force helps the motion.
Work done is zero in two cases: when there is no displacement (\(S = 0\)), and when the displacement is normal to the force (\(\theta = 90^\circ\), since \(\cos 90^\circ = 0\)).
Work done is negative when \(\theta = 180^\circ\), so that \(W = -FS\); the force opposes the motion, as friction does.
The work done by the force of gravity when a body of mass \(m\) descends through a vertical height \(h\) is \[W = mgh\] with \(m\) in kilogram (kg), \(g\) in \(\text{m s}^{-2}\) and \(h\) in metre (m). The work done by gravity is \(-mgh\) when the body rises through \(h\), and the body then does \(+mgh\) of work against gravity.
The work done by gravity depends only on the vertical height, not on the path taken. For a staircase, \(h\) is the number of steps multiplied by the height of each step.
The area enclosed between a force–displacement graph and the displacement axis gives the work done. This method works for a variable force as well as a constant one.
The S.I. unit of work is the joule (J): one joule is the work done when a force of one newton displaces a body through one metre in the direction of the force. The C.G.S. unit is the erg, where \(1\ \text{erg} = 1\ \text{dyne} \times 1\ \text{cm}\), and \[1\ \text{J} = 10^{7}\ \text{erg}\]
Bigger units of work are the kilo-joule (\(10^{3}\ \text{J}\)), the mega-joule (\(10^{6}\ \text{J}\)) and the giga-joule (\(10^{9}\ \text{J}\)).
Power is the rate of doing work: \[P = \frac{W}{t}\] where \(W\) is the work done in joule (J), \(t\) is the time taken in second (s) and \(P\) is the power in watt (W). Power is a scalar quantity.
For a constant force acting along the motion, \[P = F \times v\] where \(v\) is the average speed in \(\text{m s}^{-1}\).
The S.I. unit of power is the watt (W): one watt is the power of a source that does one joule of work in one second, so \(1\ \text{W} = 1\ \text{J s}^{-1}\). Bigger units are the kilowatt (\(10^{3}\ \text{W}\)), the megawatt (\(10^{6}\ \text{W}\)) and the gigawatt (\(10^{9}\ \text{W}\)); smaller units are the milliwatt (\(10^{-3}\ \text{W}\)) and the microwatt (\(10^{-6}\ \text{W}\)).
The C.G.S. unit of power is the erg per second, and \(1\ \text{W} = 10^{7}\ \text{erg s}^{-1}\). Another unit of power is the horse power, where \[1\ \text{H.P.} = 746\ \text{W} = 0.746\ \text{kW}\]
The energy of a body is its capacity to do work. It is a scalar quantity, measured in the same units as work. Whenever work is done, energy is transferred: a body doing work loses energy, and a body on which work is done gains energy.
Bigger units of energy are the watt hour, \(1\ \text{Wh} = 3600\ \text{J} = 3.6\ \text{kJ}\), and the kilowatt hour, \(1\ \text{kWh} = 3.6 \times 10^{6}\ \text{J}\).
Heat energy is measured in calorie: one calorie is the heat energy needed to raise the temperature of one gram of water from \(14.5^\circ\text{C}\) to \(15.5^\circ\text{C}\). \(1\ \text{J} = 0.24\ \text{calorie}\), \(1\ \text{calorie} = 4.18\ \text{J}\), and \(1\ \text{kilocalorie} = 1000\ \text{calorie} = 4180\ \text{J}\).
Energy transfers of atomic particles are measured in electron volt: one electron volt is the energy gained by an electron accelerated through a potential difference of one volt, and \[1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}\]
Mechanical energy is the energy of a body due to its state of rest or of motion. It exists in two forms, potential energy and kinetic energy, and the total mechanical energy of a body is their sum.
Potential energy \(U\) is the energy of a body at rest due to its position, or due to its changed size and shape. Gravitational potential energy is due to a changed position; elastic potential energy is due to a changed size and shape, and equals the work done in deforming the body.
The gravitational potential energy of a body is zero at infinity, and at a finite distance from the Earth it is negative, because the work of bringing the body in is done by the force of attraction of the Earth itself.
Taking the potential energy on the ground as zero, the gravitational potential energy of a body at a height \(h\) is \[U = mgh\] with \(U\) in joule (J), \(m\) in kilogram (kg), \(g\) in \(\text{m s}^{-2}\) and \(h\) in metre (m). Between two heights, the gain in potential energy is \(mg(h_2 - h_1)\) and the loss is \(mg(h_1 - h_2)\).
Kinetic energy \(K\) is the energy of a body due to its motion, and equals the work the body can do before coming to rest: \[K = \tfrac{1}{2}mv^{2}\] with \(m\) in kilogram (kg) and \(v\) in \(\text{m s}^{-1}\). Since the velocity is squared, doubling the velocity makes the kinetic energy four times as great.
Kinetic energy and momentum \(p\), in \(\text{kg m s}^{-1}\), are related by \[K = \frac{p^{2}}{2m} \qquad \text{and} \qquad p = \sqrt{2mK}\] Of two bodies with the same momentum, the lighter has the greater kinetic energy; of two bodies with the same kinetic energy, the heavier has the greater momentum.
By the work–energy theorem, the increase in the kinetic energy of a moving body equals the work done on it by a force acting in the direction of its motion: \[W = \tfrac{1}{2}mv^{2} - \tfrac{1}{2}mu^{2}\] where \(u\) and \(v\) are the initial and final velocities in \(\text{m s}^{-1}\).
Kinetic energy has three forms — translational, rotational and vibrational — and a body may possess more than one of them at the same time.
Potential energy changes into kinetic energy whenever it is put to use, as in a falling weight, an unwinding watch spring, a released bow and an expanding compressed spring.
Energy occurs in many forms: solar, heat, light, chemical, hydro, electrical, nuclear, geo-thermal, wind, sound, magnetic and mechanical energy.
One form of energy can be converted into another, and devices exist for the common conversions in both directions — a dynamo and a motor, an electric iron and a thermocouple, a loudspeaker and a microphone, a lighted bulb and a photocell.
Whenever mechanical energy changes into another form, it does so as kinetic energy; stored potential energy must first become kinetic energy.
In every conversion a part of the energy changes into an undesirable form, usually heat due to friction, or is lost to the surroundings. This is called dissipation of energy, and that part is called degraded energy. It is not destroyed, only put beyond use.
By the principle of conservation of energy, energy can neither be created nor destroyed; it only changes from one form into another, and the total energy always remains the same.
If only potential energy and kinetic energy are interchanged, the total mechanical energy stays constant: \[K + U = \text{constant}\] This holds only in the absence of all kinds of frictional forces.
For a body of mass \(m\) falling freely from a height \(h\), the total energy is \(mgh\) at the top, \(mgx + mg(h - x) = mgh\) after falling a distance \(x\), and \(mgh\) on reaching the ground. As the body falls, its potential energy decreases and its kinetic energy increases by the same amount.
The velocity of a freely falling body on reaching the ground is \(v = \sqrt{2gh}\), and the velocity of projection needed to rise to a height \(h\) is \(u = \sqrt{2gh}\), since \(\tfrac{1}{2}mu^{2} = mgh\). Neither depends on the mass of the body.
In an oscillating simple pendulum, the bob has only potential energy at the extreme positions, only kinetic energy at the mean position, and both in between, the sum remaining constant. This is strictly true only in the absence of air friction.
Flowchart
FC 2.5 — The Energy Story Of A Falling Body From Start To Finish Work done in lifting the body → gravitational potential energy stored at the height → potential energy falling and kinetic energy rising, their sum staying constant during the fall → whole of the energy as kinetic energy on reaching the ground → heat and sound energy on striking the ground, the total energy still unchanged
2.17 Key Terms
Calorie — the heat energy required to raise the temperature of one gram of water from \(14.5^\circ\text{C}\) to \(15.5^\circ\text{C}\); \(1\ \text{calorie} = 4.18\ \text{J}\).
Chemical energy — the energy stored in fuels, food and cells, which is released when they burn or react; unit: joule (J).
Electrical energy — the energy possessed by charges, released when they move, as in a rubbed body or an electric cell; unit: joule (J).
Electron volt — the energy gained by an electron accelerated through a potential difference of one volt; \(1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}\).
Energy — the capacity of a body to do work, measured by the amount of work the body can perform; S.I. unit: joule (J).
Erg — the C.G.S. unit of work and energy; the work done when a force of one dyne displaces a body through one centimetre in its own direction; \(1\ \text{J} = 10^{7}\ \text{erg}\).
Geo-thermal energy — the energy stored deep inside the Earth, released in nuclear disintegrations there and reaching the surface as natural steam or hot springs; unit: joule (J).
Heat energy — the energy released on burning fuels, and the energy that flows from a hotter body to a colder one; unit: joule (J).
Horse power — a unit of power used in mechanical engineering; \(1\ \text{H.P.} = 746\ \text{W} = 0.746\ \text{kW}\).
Hydro energy — the energy possessed by fast moving water, used to drive the turbines of a hydroelectric power station; unit: joule (J).
Joule — the S.I. unit of work and energy; the work done when a force of one newton displaces a body through one metre in the direction of the force.
Kilowatt hour — the energy spent by a source of power one kilowatt working for one hour; \(1\ \text{kWh} = 3.6 \times 10^{6}\ \text{J}\).
Kinetic energy — the energy possessed by a body due to its state of motion, equal to \(\tfrac{1}{2}mv^{2}\); S.I. unit: joule (J).
Light energy — the form of energy in the presence of which other objects are seen; unit: joule (J).
Magnetic energy — the energy possessed by a magnet, by virtue of which it can attract iron filings; unit: joule (J).
Mechanical energy — the energy possessed by a body due to its state of rest or of motion, being the sum of its potential energy and kinetic energy; S.I. unit: joule (J).
Nuclear energy — the energy released due to the loss of mass in nuclear fission and fusion, in accordance with \(E = mc^{2}\); unit: joule (J).
Potential energy — the energy possessed by a body at rest due to its position, or due to its changed size and shape; S.I. unit: joule (J).
Power — the rate of doing work, that is, the work done by a source in one second; S.I. unit: watt (W).
Solar energy — the energy radiated out by the Sun; unit: joule (J).
Sound energy — the energy carried by the disturbance produced by a vibrating body, sensed by the ear; unit: joule (J).
Watt — the S.I. unit of power; the power of a source that does one joule of work in one second, so \(1\ \text{W} = 1\ \text{J s}^{-1}\).
Watt hour — the energy spent by a source of power one watt working for one hour; \(1\ \text{Wh} = 3600\ \text{J}\).
Wind energy — the energy possessed by fast moving air, used to drive wind mills and wind generators; unit: joule (J).
Work — the product of the force applied on a body and the displacement produced in the direction of the force; work is done only when the body moves; S.I. unit: joule (J).
2.18 Points To Ponder
A porter who walks a long stretch of level ground with a trunk on his head does no work against gravity, yet he becomes hungry and tired. If none of the food energy went into work on the trunk, what happened to it, and how might you set about measuring where it went?
If energy can neither be created nor destroyed, why do people speak of an energy crisis and of saving energy? What exactly is it that a country can run short of, and what would it mean to waste something that cannot be destroyed?
The kinetic energy of a vehicle depends on the square of its speed, and all of that energy must be removed by the brakes before it stops. What does that single fact suggest about speed limits outside a school, and about why a loaded lorry needs a longer stopping distance than a car at the same speed?
Section 2.7 says the gravitational potential energy of a body at any finite distance from the Earth is negative, while section 2.8 writes \(U = mgh\), a positive quantity. Are these two statements in conflict, or can both be true at once? What is doing the work in each case?
A ball dropped on a floor rebounds to a lower height each time until it finally stops. Suppose you wanted to prove that not a single joule had been destroyed. What quantities would you have to measure, and which of them would be hardest to measure in a classroom?
Two identical pendulums are set swinging through the same displacement, one on the Earth and one on the Moon, where \(g\) is smaller. Would the two bobs carry the same total energy? Would they pass through the mean position at the same speed? Explain your reasoning from \(K + U = \text{constant}\).
2.19 Objective Questions
Multiple Choice
- (MCQ) The work done by a force on a body is maximum when the angle between the force and the displacement is: (1 mark)
- \(0^\circ\)
- \(45^\circ\)
- \(90^\circ\)
- \(180^\circ\)
- (MCQ) A stone tied to a string is whirled in a horizontal circle at a steady speed. The work done on the stone by the force directed towards the centre is: (1 mark)
- positive
- negative
- zero
- equal to its kinetic energy
- (MCQ) Which of the following is a unit of energy and not of power? (1 mark)
- watt
- kilowatt
- kilowatt hour
- horse power
- (MCQ) The velocity of a moving body is made three times, its mass remaining the same. Its kinetic energy becomes: (1 mark)
- three times
- six times
- nine times
- one-third
- (MCQ) A body of mass \(m\) is thrown vertically upwards and rises through a height \(h\). The work done on it by the force of gravity is: (1 mark)
- \(mgh\)
- \(-mgh\)
- zero
- \(\tfrac{1}{2}mgh\)
- (MCQ) One horse power is equal to: (1 mark)
- \(746\ \text{W}\)
- \(1000\ \text{W}\)
- \(3600\ \text{W}\)
- \(0.746\ \text{W}\)
Assertion–Reason
For each of the following, choose from: a. Both A and R are true and R is the correct explanation of A b. Both A and R are true but R is not the correct explanation of A c. A is false but R is true d. A is true but R is false
(Assertion–Reason) Assertion (A): The work done by the force of friction on a moving body is negative. Reason (R): The force of friction acts opposite to the direction of motion of the body. (1 mark)
(Assertion–Reason) Assertion (A): The mechanical energy of a pendulum swinging in air is not conserved. Reason (R): Energy is destroyed by the force of friction due to air. (1 mark)
(Assertion–Reason) Assertion (A): Of two bodies having the same momentum, the lighter one has the greater kinetic energy. Reason (R): For a given momentum, the kinetic energy is inversely proportional to the mass. (1 mark)
Match The Following
- (Match) Match each device in Column A with the energy conversion it brings about in Column B. (1 mark)
| Column A | Column B |
|---|---|
| a. dynamo | i. electrical energy into heat energy |
| b. microphone | ii. mechanical energy into electrical energy |
| c. electric iron | iii. light energy into electrical energy |
| d. solar cell | iv. sound energy into electrical energy |
Fill In The Blanks
(Fill in the blank) $1 = $ __________ \(\text{J}\). (1 mark)
(Fill in the blank) The area enclosed between a force–displacement graph and the displacement axis gives the __________ done by the force. (1 mark)
(Fill in the blank) The energy possessed by a body due to its changed size and shape is called __________ potential energy. (1 mark)
True Or False
(True/False) Work is a vector quantity, because force and displacement are both vectors. (1 mark)
(True/False) The total mechanical energy of a body is conserved even when friction acts on it. (1 mark)
Name The Following
(Name the following) Name the physical quantity that is measured in electron volt, and name the form of energy possessed by a wound-up watch spring. (1 mark)
(Give one word or phrase) Give one word for: the rate at which work is done by a source. (1 mark)
(State the S.I. unit) State the S.I. unit of (i) work, and (ii) power. (1 mark)
(Odd one out) Pick the odd one out and give a reason: joule, erg, watt, kilowatt hour. (1 mark)
2.20 Check Your Concepts
Define work, and state the condition under which a force does no work on a body although the force is acting on it. (2 marks)
Define power and state its S.I. unit. How is the horse power related to that unit? (2 marks)
Give a reason: a porter carrying a load on his head along level ground does no work against the force of gravity. (2 marks)
Differentiate between work and power, stating the basis of difference in each case. (2 marks)
A body slides down a slope and reaches the bottom, and an identical body is dropped straight down through the same vertical height. Give a reason why the work done by the force of gravity is the same in both cases. (3 marks)
State the work–energy theorem and write the relation that expresses it, naming each symbol with its unit. (3 marks)
Differentiate between potential energy and kinetic energy, stating any three bases of difference. (3 marks)
What is meant by degradation of energy? Explain it with one example taken from daily life, and state whether any energy is destroyed in the process. (3 marks)
State the principle of conservation of energy. Under what condition alone does the mechanical energy of a body remain constant? (4 marks)
Why does the kinetic energy of a car become four times as great when its speed is doubled, while its momentum becomes only twice as great? Explain from the two relations concerned. (4 marks)
Obtain the expression \(U = mgh\) for the gravitational potential energy of a body of mass \(m\) raised to a height \(h\) above the ground, stating clearly the assumption made about the level of zero potential energy. (5 marks)
Derive the expression \(K = \tfrac{1}{2}mv^{2}\) for the kinetic energy of a body of mass \(m\) moving with velocity \(v\), stating the assumption made about the opposing force. (5 marks)
Show that the total mechanical energy of a body of mass \(m\) falling freely from a height \(h\) is the same (i) at the top, (ii) after it has fallen through a distance \(x\), and (iii) on reaching the ground. State the condition under which the result holds. (5 marks)
Explain the energy changes that take place in an oscillating simple pendulum as its bob moves from an extreme position to the mean position and on to the other extreme position. Why does a real pendulum finally come to rest? (5 marks)
2.21 Diagrams, Graphs & Constructions
- Fig. 2.21-1 shows how the force acting on a body varies with its displacement. Using the graph, find (i) the work done during the first \(4\ \text{m}\), (ii) the work done between \(4\ \text{m}\) and \(8\ \text{m}\), and (iii) the total work done over the whole \(8\ \text{m}\). Show how you obtain each answer from the graph. (4 marks)
Diagram
Fig. 2.21-1 — Force–Displacement Graph For The Exercise Graph with the Y-axis labelled ‘force F (N)’, marked in steps of 5 up to 20, and the X-axis labelled ‘displacement S (m)’, marked in steps of 2 up to 8. Draw a straight line from the origin rising to the point (4, 15), then a horizontal line from (4, 15) to (8, 15). Drop dashed vertical lines from (4, 15) and (8, 15) to the X-axis, and draw a dashed horizontal line from 15 on the Y-axis across to the plotted line.
Sketch a graph showing how the kinetic energy and the potential energy of a body falling freely from a height \(h\) vary with its height above the ground. Label both axes with their units, mark both lines, and mark the height at which the two energies are equal. (4 marks)
Draw a labelled diagram of a simple pendulum displaced to one side, marking the point of suspension, the mean position and the two extreme positions. Against each of the three positions of the bob, state the form of mechanical energy it possesses there. (3 marks)
Draw a flow diagram showing, in order, the energy conversions that take place from the water stored high in the reservoir of a dam to the electrical energy leaving the power station. (3 marks)
Fig. 2.21-2 shows a block being pulled at a steady speed up a slope by a force applied along the slope. Copy the figure, mark on it the vertical height gained by the block and the distance moved along the slope, and state which of the two is used in calculating the increase in the potential energy of the block. (3 marks)
Diagram
Fig. 2.21-2 — Block Pulled Up A Slope A right-angled triangle with the horizontal base at the bottom, the right angle at the lower right corner, and the sloping side rising from the lower left corner A to the upper right corner B. Draw a small shaded block on the slope near A, with an arrow along the slope pointing up towards B labelled F. Mark the sloping side AB as 5 m. Mark the vertical side at the right of the triangle as 2 m, leaving both marks unlabelled otherwise.
2.22 Practice With Numericals
A constant force of \(25\ \text{N}\) displaces a body through \(4\ \text{m}\) in its own direction. Calculate the work done by the force. (2 marks)
A body of mass \(5\ \text{kg}\) is raised through a vertical height of \(3\ \text{m}\). Taking \(g = 10\ \text{m s}^{-2}\), calculate the work done against the force of gravity. (2 marks)
A machine does \(6000\ \text{J}\) of work in \(20\ \text{s}\). Calculate the power of the machine. (2 marks)
Calculate the kinetic energy of a body of mass \(8\ \text{kg}\) moving with a velocity of \(3\ \text{m s}^{-1}\). (2 marks)
Express \(3.5 \times 10^{7}\ \text{erg}\) in joule, and express a power of \(1.5\ \text{kW}\) in horse power. (2 marks)
A force of \(40\ \text{N}\) pulls a box through a displacement of \(10\ \text{m}\), the force making an angle of \(60^\circ\) with the displacement. Calculate the work done by the force. (3 marks)
A body of mass \(2\ \text{kg}\) has a momentum of \(12\ \text{kg m s}^{-1}\). Calculate (i) its kinetic energy, and (ii) its velocity. (3 marks)
A pump raises \(250\ \text{kg}\) of water through a vertical height of \(12\ \text{m}\) in \(40\ \text{s}\). Taking \(g = 10\ \text{m s}^{-2}\), calculate (i) the work done by the pump, and (ii) the power at which it works. (3 marks)
An electric motor rated \(2\ \text{kW}\) is used for \(5\ \text{h}\). Calculate the energy it consumes, expressed in (i) kilowatt hour, and (ii) joule. (3 marks)
A trolley of mass \(3\ \text{kg}\) is pushed along a level floor and its velocity increases from \(2\ \text{m s}^{-1}\) to \(6\ \text{m s}^{-1}\). Calculate the work done on the trolley by the pushing force. (3 marks)
A boy of mass \(40\ \text{kg}\) climbs a flight of 30 steps, each \(20\ \text{cm}\) high, in \(50\ \text{s}\). Taking \(g = 10\ \text{m s}^{-2}\), calculate (i) the work done by him against the force of gravity, and (ii) the power he spends. (4 marks)
A stone of mass \(0.4\ \text{kg}\) is dropped from a height of \(20\ \text{m}\) above the ground. Taking \(g = 10\ \text{m s}^{-2}\) and neglecting air friction, calculate (i) its potential energy at the top, (ii) its kinetic energy just before striking the ground, and (iii) the velocity with which it strikes the ground. (4 marks)
A ball of mass \(0.1\ \text{kg}\) is thrown vertically upwards with a velocity of \(20\ \text{m s}^{-1}\). Taking \(g = 10\ \text{m s}^{-2}\) and neglecting air friction, calculate (i) the initial kinetic energy imparted to the ball, (ii) the greatest height it reaches, and (iii) its potential energy at that height. (4 marks)
A locomotive develops a power of \(60\ \text{kW}\) while pulling a train at a constant speed against a total resistive force of \(4000\ \text{N}\). Calculate the speed of the train. (4 marks)
A car of mass \(1000\ \text{kg}\) increases its speed from \(36\ \text{km h}^{-1}\) to \(72\ \text{km h}^{-1}\) on a level road. Calculate the work done by the engine of the car in doing so. (5 marks)