5.1 Introduction
Chapter 5 — Work, Energy and Power
In the 1840s, a young British brewer and physicist named James Prescott Joule became fascinated by a simple question: what happens to the energy of a moving object when it slows down and stops? Many scientists of his time believed heat was a kind of invisible fluid that flowed in and out of matter. Joule suspected something bolder — that motion itself could be turned into heat, and that the two were really the same thing, just measured in different ways.
To test this, he built a clever apparatus. A slowly falling weight turned a paddle wheel sealed inside a container of water. As the paddles churned against the water, Joule measured the water’s temperature with extraordinary care. The falling weight did mechanical “work”; the water grew measurably warmer. Repeating the experiment again and again, he showed that a fixed amount of work always produced the same fixed amount of heating.
This was a turning point in physics. Joule had shown that energy is not destroyed when motion stops — it is only converted from one form into another. His painstaking measurements helped establish the conservation of energy, one of the deepest principles in all of science. Today the SI unit of work and energy, the joule, carries his name.
Figure to come
Fig. 5.0 – Joule’s paddle-wheel apparatus: a falling weight connected by a cord over a pulley, turning paddles inside an insulated water container fitted with a thermometer.
By the end of this chapter, you will be able to answer each of these questions using the ideas of work, energy, and power.
In everyday language, the words work, energy, and power are used freely and often loosely. A farmer ploughing a field, a labourer carrying bricks up a ladder, a student preparing for an exam, an artist painting a landscape — we say all of them are “working.”
In physics, however, the word work is given a sharp, precise meaning. It is not enough to feel tired or to be busy; whether “work” has been done depends on a definite condition that we will build up carefully in this chapter.
The word energy is tied to work even in daily speech. When we say someone has the stamina to work fourteen hours a day, we mean they have a large capacity to do work. This is close to the physics idea too: energy is the capacity to do work. But because work is defined precisely in physics, energy also becomes a precise, measurable quantity.
The word power also carries everyday shades of meaning. A “powerful” punch in boxing is one delivered quickly and forcefully. This hints at the physical idea, where power measures how fast work is done — but again, physics fixes an exact definition.
So there is only a loose match between these physical quantities and the mental pictures we attach to the same words. The aim of this chapter is to replace those vague pictures with clear, usable definitions of work, energy, and power.
Before we can do that, we need one mathematical tool: a way to “multiply” two vectors that gives us a single number. This tool is the scalar product, and we develop it first.
Figure to come
Fig. 5.1(a) – Two vectors A and B drawn from a common origin with the angle θ between them marked.
5.1.1 The Scalar Product
In an earlier chapter you learned that many physical quantities — displacement, velocity, acceleration, force — are vectors, meaning they have both magnitude and direction. You also learned to add and subtract vectors.
We now need to multiply vectors. There are two different ways to do this. One way, the scalar product, combines two vectors to give a single scalar (a plain number). The other, the vector product, combines two vectors to give a new vector; we will study it in a later chapter. Here we focus only on the scalar product.
The scalar product (also called the dot product) of two vectors \(\vec{A}\) and \(\vec{B}\), written \(\vec{A} \cdot \vec{B}\) (read “A dot B”), is defined as
\[\vec{A} \cdot \vec{B} = A\,B\cos\theta \qquad (5.1a)\]
Here \(A\) and \(B\) are the magnitudes of the two vectors and \(\theta\) is the angle between them (both drawn from a common starting point, as in Fig. 5.1(a)). All three of \(A\), \(B\), and \(\cos\theta\) are just numbers, so their product is a plain number.
This is worth pausing on. Each of \(\vec{A}\) and \(\vec{B}\) points in some direction, yet their dot product \(\vec{A}\cdot\vec{B}\) has no direction at all. Multiplying two vectors this way “throws away” direction and leaves only a number.
The geometric meaning: projection. We can rewrite Eq. (5.1a) in two ways by grouping the terms differently:
\[\vec{A} \cdot \vec{B} = A\,(B\cos\theta) = B\,(A\cos\theta)\]
The quantity \(B\cos\theta\) is the length of \(\vec{B}\)’s “shadow” cast along the direction of \(\vec{A}\) — this is called the projection of \(\vec{B}\) onto \(\vec{A}\), shown in Fig. 5.1(b). Likewise, \(A\cos\theta\) is the projection of \(\vec{A}\) onto \(\vec{B}\), shown in Fig. 5.1(c).
Figure to come
Fig. 5.1(b) – Vector B with a dashed perpendicular dropped onto the direction of A, showing the projection length B cos θ along A.
Figure to come
Fig. 5.1(c) – Vector A with a dashed perpendicular dropped onto the direction of B, showing the projection length A cos θ along B.
So the dot product can be read as: the magnitude of one vector times the component of the other vector along it. This “component along a direction” idea is exactly what we will need when we define work.
Properties of the scalar product. The dot product obeys the commutative law, meaning the order does not matter:
\[\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}\]
It also obeys the distributive law over addition:
\[\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}\]
And if \(\lambda\) is any real number,
\[\vec{A} \cdot (\lambda \vec{B}) = \lambda\,(\vec{A} \cdot \vec{B})\]
Dot products of unit vectors. Recall the unit vectors \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) that point along the \(x\)-, \(y\)-, and \(z\)-axes. Each has magnitude 1, and the axes are mutually perpendicular. Using the definition with \(\cos 0^\circ = 1\) (a vector with itself) and \(\cos 90^\circ = 0\) (two different axes), we get
\[\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1\]
\[\hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0\]
Component form. Any two vectors can be written in terms of their components:
\[\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}, \qquad \vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}\]
Multiplying these out term by term and using the unit-vector results above (all the “mixed” terms vanish because they contain a zero dot product), the scalar product simplifies to
\[\vec{A}\cdot\vec{B} = A_xB_x + A_yB_y + A_zB_z \qquad (5.1b)\]
This is the form most useful for calculation: just multiply matching components and add. Here \(A_x, A_y, A_z\) and \(B_x, B_y, B_z\) are the components of the two vectors along the three axes (each in the same units as the vector itself).
Two special results follow directly from Eq. (5.1b):
(i) The dot product of a vector with itself gives the square of its magnitude:
\[\vec{A}\cdot\vec{A} = A_xA_x + A_yA_y + A_zA_z\]
\[A^2 = A_x^2 + A_y^2 + A_z^2 \qquad (5.1c)\]
This makes sense because \(\vec{A}\cdot\vec{A} = |\vec{A}||\vec{A}|\cos 0^\circ = A^2\). It is just the Pythagoras relation for the magnitude of a vector.
(ii) The dot product is zero when the two vectors are perpendicular, since \(\cos 90^\circ = 0\). This gives a quick test: if \(\vec{A}\cdot\vec{B} = 0\) (and neither vector is zero), the vectors are at right angles.
This projection idea — the component of a force along a displacement — is precisely what we will use to define work in Section 5.3.
5.2 Notions of Work and Kinetic Energy: The Work-Energy Theorem
We now have the mathematical tool we need — the scalar product. Let us use it to arrive naturally at two of the central ideas of this chapter: kinetic energy and work. Rather than defining them out of the blue, we will let them emerge from a piece of kinematics you already know.
Recall the equation of motion for a body moving in a straight line with constant acceleration \(a\):
\[v^2 - u^2 = 2as \qquad (5.2)\]
Here \(u\) is the initial speed, \(v\) is the final speed (both in \(\text{m s}^{-1}\)), \(a\) is the constant acceleration (\(\text{m s}^{-2}\)), and \(s\) is the distance travelled (m). This relation was established back in the study of motion; here we simply put it to new use.
Multiply both sides of Eq. (5.2) by \(m/2\), where \(m\) is the mass of the body (kg):
\[\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mas\]
Now recall Newton’s second law, \(F = ma\). So the product \(mas\) can be written as \((ma)s = Fs\), where \(F\) is the net force. This gives
\[\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mas = Fs \qquad (5.2a)\]
So far this is one-dimensional. Motion in general is three-dimensional, so we rewrite the same relation using vectors. The kinematic result becomes
\[v^2 - u^2 = 2\,\vec{a}\cdot\vec{d}\]
where \(\vec{a}\) is the acceleration vector and \(\vec{d}\) is the displacement vector of the object. Notice that the scalar product from Section 5.1 has appeared exactly as promised — it is what lets us combine the two vectors into a single number.
Multiplying both sides again by \(m/2\), and using Newton’s second law in vector form \(\vec{F} = m\vec{a}\):
\[\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = m\,\vec{a}\cdot\vec{d} = \vec{F}\cdot\vec{d} \qquad (5.2b)\]
Figure to come
Fig. 5.2A – A block of mass m speeding up from initial speed u to final speed v as a net force F pushes it through displacement d.
This equation is the doorway to two definitions. Look at each side separately.
The left side is the change in the quantity “half the mass times the square of the speed,” from its initial value to its final value. We give this quantity a name: the kinetic energy of the body, denoted \(K\). So the left side is simply \(K_f - K_i\), the final kinetic energy minus the initial kinetic energy.
The right side, \(\vec{F}\cdot\vec{d}\), is the product of the displacement and the component of the force along that displacement — exactly the projection idea from the scalar product. This quantity is called the work done by the force, denoted \(W\).
With these two names, Eq. (5.2b) reads
\[K_f - K_i = W \qquad (5.3)\]
where \(K_i\) and \(K_f\) are the initial and final kinetic energies of the object.
This is the central relation. Work refers to a force acting through a displacement — that is, work is done by a force on a body over a certain displacement. And Eq. (5.3) tells us that this work shows up as a change in the body’s kinetic energy.
Equation (5.2) that we started from is just a special case — for a constant force in a straight line — of a much more general result called the work-energy theorem.
We derived this theorem here for a constant force. Later, in Section 5.6, we will show it holds even when the force varies from point to point.
5.3 Work
In the last section, work appeared naturally as the quantity \(\vec{F}\cdot\vec{d}\). Let us now look at it directly and understand exactly what it measures.
As we saw, work is tied to two things: a force and the displacement over which that force acts. Consider a constant force \(\vec{F}\) acting on an object of mass \(m\). The object moves through a displacement \(\vec{d}\) in the positive \(x\)-direction, while the force points at an angle \(\theta\) to that displacement, as shown in Fig. 5.2.
Figure to come
Fig. 5.2 – A block on a surface undergoing displacement d in the x-direction while a constant force F acts on it at angle θ above the horizontal.
Only the part of the force that lies along the displacement does work. That part is the component \(F\cos\theta\) (the projection of the force onto the direction of motion). Multiplying it by the distance moved gives the work.
\[W = (F\cos\theta)\,d = \vec{F}\cdot\vec{d} \qquad (5.4)\]
Here \(F\) is the magnitude of the force (N), \(d\) is the magnitude of the displacement (m), \(\theta\) is the angle between them, and \(W\) is the work done (joule, J).
Work in physics is not the same as “effort.” This is the point where physics parts ways with everyday language. If there is no displacement, there is no work — no matter how large the force.
Push as hard as you like against a rigid brick wall: the wall does not move, so the work you do on it is zero. Yet your muscles keep contracting and relaxing, internal energy is spent, and you tire out. Feeling tired is not the same as doing work in the physics sense.
When is no work done? Equation (5.4) tells us the work is zero in three distinct situations.
(i) The displacement is zero. This is the wall case. A weightlifter holding a 150 kg barbell steadily on their shoulders for 30 s does no work on the load during that time, because the load does not move.
(ii) The force is zero. A block sliding on a perfectly smooth (frictionless) horizontal table feels no horizontal force, yet it can move a large distance. With no force, there is no work — even with plenty of displacement.
(iii) The force and displacement are perpendicular. When \(\theta = \pi/2\) rad \((= 90^\circ)\), \(\cos(\pi/2) = 0\), so \(W = 0\). For the block moving on the smooth table, gravity \(mg\) pulls straight down while the motion is horizontal — so gravity does no work on it.
The same idea applies on a grand scale. If we assume the Moon’s orbit around the Earth is a perfect circle, the Earth’s gravitational pull on the Moon does no work. At every instant the Moon’s displacement is along the circle (tangential), while the Earth’s force points inward toward the centre (radial), so the two are at right angles, \(\theta = \pi/2\).
Work can be positive or negative. The sign comes entirely from \(\cos\theta\) in Eq. (5.4).
When \(\theta\) lies between \(0^\circ\) and \(90^\circ\), \(\cos\theta\) is positive, so the work is positive — the force helps the motion.
When \(\theta\) lies between \(90^\circ\) and \(180^\circ\), \(\cos\theta\) is negative, so the work is negative — the force opposes the motion. A very common case is friction, which acts opposite to the displacement, so \(\theta = 180^\circ\) and \(\cos 180^\circ = -1\). Work done by friction is therefore negative.
Figure to come
Fig. 5.2B – Two panels: (left) force at an acute angle to displacement giving positive work; (right) force at an obtuse angle to displacement giving negative work.
Units and dimensions. From Eq. (5.4), work is force times distance, so its dimensions are \([\text{ML}^2\text{T}^{-2}]\) — the same as energy. The SI unit is the joule (J).
Because work and energy appear across so many fields, several other units are also in common use. Some of these are listed in Table 5.1.
Table 5.1 — Alternative Units of Work / Energy in J
| Unit | Value in joule |
|---|---|
| erg | \(10^{-7}\) J |
| electron volt (eV) | \(1.6\times10^{-19}\) J |
| calorie (cal) | \(4.186\) J |
| kilowatt hour (kWh) | \(3.6\times10^{6}\) J |
5.4 Kinetic Energy
We met kinetic energy in Section 5.2 as one side of the work-energy theorem. Let us now write it down properly and understand what it tells us.
If an object of mass \(m\) moves with velocity \(\vec{v}\), its kinetic energy \(K\) is
\[K = \frac{1}{2}m\,\vec{v}\cdot\vec{v} = \frac{1}{2}mv^2 \qquad (5.5)\]
The middle step uses the scalar product from Section 5.1: dotting a vector with itself gives the square of its magnitude, \(\vec{v}\cdot\vec{v} = v^2\). So the compact vector form and the familiar \(\tfrac{1}{2}mv^2\) are the same thing. Here \(m\) is the mass (kg), \(v\) is the speed (\(\text{m s}^{-1}\)), and \(K\) is in joule (J).
Because \(\vec{v}\cdot\vec{v}\) depends only on the magnitude of velocity, kinetic energy is a scalar — it has no direction. A body moving north and a body moving south at the same speed have exactly the same kinetic energy.
Physically, the kinetic energy of an object measures the work it can do by virtue of its motion. A moving object can push, break, or drive other things — and how much it can do is captured by \(K\).
This idea has been used by people for a very long time, well before it was written as a formula. The kinetic energy of a fast-flowing stream was harnessed to turn wheels and grind corn, and sailing ships have always used the kinetic energy of the wind to move.
Table 5.2 lists the kinetic energies of some everyday objects. Notice how enormously they differ — from a whole car down to a single air molecule.
Table 5.2 — Typical kinetic energies (K)
| Object | Mass (kg) | Speed (m s⁻¹) | K (J) |
|---|---|---|---|
| Car | 2000 | 25 | \(6.3\times10^{5}\) |
| Running athlete | 70 | 10 | \(3.5\times10^{3}\) |
| Bullet | \(5\times10^{-2}\) | 200 | \(10^{3}\) |
| Stone dropped from 10 m | 1 | 14 | \(10^{2}\) |
| Rain drop at terminal speed | \(3.5\times10^{-5}\) | 9 | \(1.4\times10^{-3}\) |
| Air molecule | \(\simeq 10^{-26}\) | 500 | \(\simeq 10^{-21}\) |
This single quantity, kinetic energy, lets us place wildly different objects on one common scale. A raindrop and a bullet are utterly different in size and speed, yet Table 5.2 puts a number to the “amount of motion” each carries — about \(1.4\times10^{-3}\) J for the raindrop and about \(10^{3}\) J for the bullet. That is why the bullet can shatter wood while the raindrop cannot: their kinetic energies differ by roughly a million times.
Figure to come
Fig. 5.4A – A parabola of kinetic energy K versus speed v, showing that K rises steeply (as v²), with dashed lines marking how doubling v quadruples K.
5.5 Work Done by a Variable Force
So far we have found the work done by a constant force using \(W = \vec{F}\cdot\vec{d}\). But in real life a truly constant force is rare. Most forces change as the object moves — they may grow, shrink, or reverse from point to point. Such a force is called a variable force, and it is what we usually meet in practice.
Consider a force that acts along one direction but whose size changes with position. If we plot this force \(F(x)\) against the position \(x\), we get a curve rather than a flat line, as shown in Fig. 5.3.
Figure to come
Fig. 5.3 – A curve of force F(x) versus position x, rising and falling smoothly across the interval from x_i to x_f.
We cannot simply multiply “force times distance” here, because the force does not have one single value over the whole path. The trick is to break the journey into pieces so small that, within each piece, the force barely changes.
Take a very small displacement \(\Delta x\). Over such a tiny step, the force \(F(x)\) stays almost constant, so the small amount of work done in that step is approximately
\[\Delta W = F(x)\,\Delta x\]
This is just “force times distance” applied to one narrow strip. On the graph, \(F(x)\,\Delta x\) is the area of a thin rectangle of height \(F(x)\) and width \(\Delta x\), as illustrated in Fig. 5.3(a).
Figure to come
Fig. 5.3(a) – The area under the F(x) curve approximated by many thin rectangles of width Δx standing side by side from x_i to x_f.
To get the total work from the start point \(x_i\) to the end point \(x_f\), we add up the work done in all these little strips:
\[W \cong \sum_{x_i}^{x_f} F(x)\,\Delta x \qquad (5.6)\]
where the sum runs from the initial position \(x_i\) to the final position \(x_f\). This is only approximate, because each rectangle has a flat top while the real curve is sloped, leaving tiny gaps and overshoots.
Now make the strips thinner and thinner. As \(\Delta x\) shrinks toward zero, the number of rectangles grows without limit, and their combined area gets closer and closer to the exact area under the curve shown in Fig. 5.3(b). In this limit, the sum becomes an integral:
\[W = \lim_{\Delta x \to 0}\sum_{x_i}^{x_f} F(x)\,\Delta x = \int_{x_i}^{x_f} F(x)\,dx \qquad (5.7)\]
Here “\(\lim\)” means the limiting value of the sum as \(\Delta x\) tends to zero. So the work done by a varying force is the definite integral of force over displacement — and geometrically, it is the area between the force curve and the \(x\)-axis, from \(x_i\) to \(x_f\).
Figure to come
Fig. 5.3(b) – The smooth F(x) curve with the whole region beneath it shaded, labelled “Work,” from x_i to x_f.
5.6 The Work-Energy Theorem for a Variable Force
In Section 5.2 we proved the work-energy theorem for a constant force. Now that we know how to handle a varying force (Section 5.5), we can show that the same theorem holds even when the force changes from point to point. For simplicity we work in one dimension.
The idea is to track how the kinetic energy changes with time. The time rate of change of kinetic energy is
\[\frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2}mv^2\right)\]
Differentiating \(\tfrac{1}{2}mv^2\) with respect to time (the mass \(m\) is constant, and \(v\) changes with time) gives
\[= m\frac{dv}{dt}\,v\]
Now bring in Newton’s second law. The net force is \(F = ma = m\dfrac{dv}{dt}\), so we can replace \(m\dfrac{dv}{dt}\) by \(F\):
\[= F\,v \qquad \text{(from Newton's Second Law)}\]
Finally, speed is the rate of change of position, \(v = \dfrac{dx}{dt}\), so
\[= F\frac{dx}{dt}\]
Comparing the first and last lines, \(\dfrac{dK}{dt} = F\dfrac{dx}{dt}\). The \(dt\) cancels, leaving a clean relation between a small change in kinetic energy and a small displacement:
\[dK = F\,dx\]
To get the total change, we integrate from the initial position \(x_i\) (where the kinetic energy is \(K_i\)) to the final position \(x_f\) (where it is \(K_f\)):
\[\int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F\,dx\]
The left side integrates simply to \(K_f - K_i\):
\[K_f - K_i = \int_{x_i}^{x_f} F\,dx \qquad (5.8a)\]
But from Eq. (5.7), the right-hand side is precisely the work \(W\) done by the variable force. Therefore
\[K_f - K_i = W \qquad (5.8b)\]
This proves the work-energy theorem for a variable force — the change in kinetic energy still equals the work done by the net force, even when that force is not constant.
A word of caution about what the theorem does and does not contain. The work-energy theorem is powerful, but it does not carry the full content of Newton’s second law.
Newton’s second law is a statement about force and acceleration at every instant of time. The work-energy theorem, by contrast, involves an integral over the whole path. In taking that integral, the moment-by-moment time information is “summed over” and is no longer visible. So the theorem tells us the net change in kinetic energy, but not when each part of it happened.
There is a second difference. Newton’s second law in two or three dimensions is a vector equation, carrying full information about directions. The work-energy theorem is a scalar equation. In going to the scalar form, the directional information in Newton’s law is not retained.
5.7 The Concept of Potential Energy
The word potential suggests possibility, or a capacity for action that is not being used just yet. In physics, potential energy is exactly that: stored energy — energy a body has because of where it is or how it is arranged, ready to be released.
A stretched bow-string is a good picture of this. While it is held drawn, it does nothing, but it holds energy. Release it, and that stored energy is handed to the arrow, which flies off at great speed.
The same idea operates on a planetary scale. The Earth’s crust is not smooth and uniform; it has cracks and dislocations called fault lines. Along these faults, rock is slowly bent and strained over long times, behaving rather like compressed springs and storing enormous amounts of potential energy.
So potential energy is the “stored energy” a body has by virtue of its position or configuration. Left to itself, the body releases this stored energy in the form of kinetic energy. Let us now make this idea precise.
Gravitational potential energy. Consider a ball of mass \(m\) near the Earth’s surface. The gravitational force on it is \(mg\), directed downward, where \(g\) is the acceleration due to gravity (\(\text{m s}^{-2}\)).
We treat \(g\) as constant here. By “near the surface” we mean that the height \(h\) of the ball above the ground is very small compared with the Earth’s radius \(R_E\) (that is, \(h \ll R_E\)), so the small change in \(g\) with height can be ignored. Throughout, we take the upward direction as positive.
Now raise the ball slowly to a height \(h\). To lift it, an external agency must push up against gravity, doing work equal to \(mgh\). This work does not vanish — it gets stored in the raised ball as gravitational potential energy.
The gravitational potential energy at height \(h\), written \(V(h)\), is defined as the negative of the work done by the gravitational force in raising the object to that height. Since gravity points down while the ball moves up, gravity does negative work \(-mgh\), and the negative of that is
\[V(h) = mgh\]
Force as the slope of potential energy. If we let \(h\) vary, the gravitational force turns out to be the negative of the rate of change of \(V(h)\) with height:
\[F = -\frac{d}{dh}V(h) = -mg\]
The negative sign tells us the gravitational force points downward — toward lower potential energy. In general, a force always points in the direction in which potential energy decreases.
Release: potential energy becomes kinetic energy. When the raised ball is let go, it falls and speeds up. Just before it hits the ground, its speed follows from the kinematic relation
\[v^2 = 2gh\]
Multiplying both sides by \(m/2\) rewrites this as
\[\frac{1}{2}mv^2 = mgh\]
The left side is the kinetic energy at the ground; the right side is the potential energy it had at height \(h\). So the gravitational potential energy stored at height \(h\) reappears, in full, as kinetic energy when the ball reaches the ground.
Figure to come
Fig. 5.7A – A ball raised to height h above the ground with potential energy mgh marked, and an arrow showing it converting to kinetic energy ½mv² as it falls.
Generalising potential energy. The idea of potential energy applies only to a special class of forces: those for which the work done against the force gets stored and can be fully recovered as kinetic energy when the constraints are removed. Such forces are called conservative forces.
Mathematically (taking one dimension for simplicity), a potential energy \(V(x)\) can be defined whenever the force \(F(x)\) can be written as
\[F(x) = -\frac{dV}{dx}\]
This relation implies, on integrating,
\[\int_{x_i}^{x_f} F(x)\,dx = -\int_{V_i}^{V_f} dV = V_i - V_f\]
So the work done by such a force depends only on the starting and ending potential energies — that is, only on the initial and final positions, not on the path taken between them.
Why path-independence matters. In earlier work with inclined planes, you saw a striking result: if an object of mass \(m\) is released from rest at the top of a smooth (frictionless) incline of height \(h\), its speed at the bottom is \(\sqrt{2gh}\) — regardless of the angle of the incline.
Figure to come
Fig. 5.7B – Two frictionless inclined planes of different angles but the same height h; a ball released from the top of each reaches the bottom with the same speed √(2gh).
So at the bottom of any such incline the object acquires the same kinetic energy, \(mgh\). This is the hallmark of a conservative force: the outcome depends only on the height dropped, not on the route. If instead the work done (or the kinetic energy gained) depended on the speed or the particular path taken, the force would be called non-conservative.
Dimensions and units. Potential energy has the same dimensions as work and kinetic energy, \([\text{ML}^2\text{T}^{-2}]\), and the same SI unit, the joule (J).
To restate the key relation: for a conservative force, the change in potential energy \(\Delta V\) equals the negative of the work done by the force:
\[\Delta V = -F(x)\,\Delta x \qquad (5.9)\]
In the falling-ball example we saw potential energy turn smoothly into kinetic energy. This points to one of the deepest ideas in mechanics — the conservation of mechanical energy — which we examine next.
5.8 The Conservation of Mechanical Energy
In the last section we saw a raised ball’s potential energy turn completely into kinetic energy as it fell. That was a glimpse of a deep and powerful principle: for certain forces, the total mechanical energy of a body stays fixed, even as it changes form. Let us establish this carefully.
For simplicity we work in one dimension. Suppose a body moves through a small displacement \(\Delta x\) under a conservative force \(F\). From the work-energy theorem, the change in kinetic energy is
\[\Delta K = F(x)\,\Delta x\]
If the force is conservative, we can define a potential energy \(V(x)\) (from Eq. 5.9) such that the work done equals the drop in potential energy:
\[-\Delta V = F(x)\,\Delta x\]
The right-hand sides of these two equations are identical. So the left-hand sides must be equal too, giving \(\Delta K = -\Delta V\), or
\[\Delta K + \Delta V = 0\]
\[\Delta(K + V) = 0 \qquad (5.10)\]
This tells us something remarkable: the change in the quantity \((K + V)\) is zero. That means \(K + V\) — the sum of the kinetic and potential energies — does not change at all. It is a constant.
Applying this over the whole path, from an initial point \(x_i\) to a final point \(x_f\):
\[K_i + V(x_i) = K_f + V(x_f) \qquad (5.11)\]
The quantity \(K + V(x)\) is called the total mechanical energy of the system. Individually, the kinetic energy \(K\) and the potential energy \(V(x)\) may vary from point to point along the path, but their sum stays the same throughout. Now the aptness of the name “conservative force” is clear — such forces conserve the total mechanical energy.
What makes a force conservative? There are three equivalent ways to recognise a conservative force, and it is worth seeing all three.
First, a force \(F(x)\) is conservative if it can be obtained from a scalar potential energy \(V(x)\) through the relation in Eq. (5.9). (The three-dimensional version of this needs a vector derivative, which is beyond this book.)
Second, the work done by a conservative force depends only on the end points of the motion — not on the path taken between them. This is clear from
\[W = K_f - K_i = V(x_i) - V(x_f)\]
which involves only the starting and ending positions.
Third, the work done by a conservative force over any closed path (one that returns to its starting point) is zero. This follows from Eq. (5.11): if \(x_i = x_f\), the start and end potential energies are equal, so the net work is zero.
These lead directly to the central principle of this section.
Making it concrete: the falling ball. Consider a ball of mass \(m\) dropped from a cliff of height \(H\), as shown in Fig. 5.5. We track its total mechanical energy at three heights: the ground (zero level), an intermediate height \(h\), and the top \(H\).
Figure to come
Fig. 5.5 – A ball of mass m dropped from a cliff of height H, marked with v = 0 at the top, v = v_h at height h, and v = v_f at the ground, showing potential energy converting to kinetic energy.
At the top, the ball is at rest, so its energy is purely potential:
\[E_H = mgH \qquad (5.11a)\]
At the intermediate height \(h\), it has both potential energy and kinetic energy (with speed \(v_h\)):
\[E_h = mgh + \frac{1}{2}mv_h^2 \qquad (5.11b)\]
At the ground, all the energy is kinetic (with final speed \(v_f\)):
\[E_0 = \frac{1}{2}mv_f^2 \qquad (5.11c)\]
A constant force such as gravity is just a special case of a position-dependent force \(F(x)\), so it is conservative and the mechanical energy is conserved. Therefore all three energies are equal.
Setting \(E_H = E_0\):
\[mgH = \frac{1}{2}mv_f^2\]
\[v_f = \sqrt{2gH}\]
This is exactly the freely-falling-body result we obtained in Section 5.7 — reached here purely from energy conservation, without using the equations of motion.
Setting \(E_H = E_h\) gives, after cancelling and rearranging,
\[v_h^2 = 2g(H - h) \qquad (5.11d)\]
which is a familiar result from kinematics.
So at height \(H\) the energy is purely potential; at height \(h\) it is part potential and part kinetic; and at the ground it is fully kinetic. The total never changes — a clean illustration of the conservation of mechanical energy.
5.9 The Potential Energy of a Spring
The spring force is a perfect example of a variable force that is also conservative. Studying it will let us put together everything from the last three sections — variable force, potential energy, and conservation of mechanical energy — in one neat package.
Consider a block resting on a smooth (frictionless) horizontal surface, attached to one end of a spring whose other end is fixed to a rigid wall, as shown in Fig. 5.7. The spring is light, so we treat it as massless.
Figure to come
Fig. 5.7 – A block attached to a spring against a wall on a smooth surface: (a) natural length, x = 0, F_s = 0; (b) stretched, x > 0, F_s < 0; (c) compressed, x < 0, F_s > 0; (d) a straight-line graph of F_s versus x with a shaded triangle showing the work done.
For an ideal spring, the spring force \(F_s\) is proportional to \(x\), the displacement of the block from its equilibrium position (the position where the spring is at its natural length). This displacement can be positive when the spring is stretched (Fig. 5.7(b)) or negative when it is compressed (Fig. 5.7(c)). The force law is called Hooke’s law:
\[F_s = -kx\]
Here \(x\) is the displacement from equilibrium (m), and \(k\) is a positive constant called the spring constant, with SI unit \(\text{N m}^{-1}\). A large \(k\) means a stiff spring (hard to stretch); a small \(k\) means a soft spring.
Work done by the spring force. Suppose we pull the block outward to an extension \(x_m\) (Fig. 5.7(b)). Since the spring force varies with position, we find the work it does using the integral for a variable force:
\[W_s = \int_0^{x_m} F_s\,dx = -\int_0^{x_m} kx\,dx = -\frac{kx_m^2}{2} \qquad (5.15)\]
The same result can be read off geometrically as the area of the shaded triangle in Fig. 5.7(d). Because \(F_s\) and \(x\) have opposite signs, this work comes out negative — the spring resists being stretched.
Meanwhile, the external force \(F\) that does the pulling does positive work, since it acts along the motion and overcomes the spring force:
\[W = +\frac{kx_m^2}{2} \qquad (5.16)\]
Compression works the same way. If instead the spring is compressed by a displacement \(x_c\) (which is negative), the spring force does work \(W_s = -kx_c^2/2\), while the external force does work \(+kx_c^2/2\) — the same pattern as for stretching.
More generally, if the block moves from an initial displacement \(x_i\) to a final displacement \(x_f\), the work done by the spring force is
\[W_s = -\int_{x_i}^{x_f} kx\,dx = \frac{kx_i^2}{2} - \frac{kx_f^2}{2} \qquad (5.17)\]
This depends only on the two end points \(x_i\) and \(x_f\) — a signature of a conservative force.
To make this vivid, consider pulling the block out to \(x_i\) and letting it return all the way back to \(x_i\) (a round trip). Then the final and initial positions are the same, and
\[W_s = -\int_{x_i}^{x_i} kx\,dx = \frac{kx_i^2}{2} - \frac{kx_i^2}{2} = 0 \qquad (5.18)\]
The work done by the spring force over a complete cycle is zero.
We have now shown two things about the spring force: (i) it depends only on position, as Hooke stated, \(F_s = -kx\); and (ii) the work it does depends only on the initial and final positions (Eq. 5.17). Both are hallmarks of a conservative force, so the spring force is conservative.
Potential energy of a spring. Since the spring force is conservative, it has a potential energy. We choose the potential energy \(V(x)\) to be zero when the block is at the equilibrium position. The analysis above then gives, for an extension or compression \(x\),
\[V(x) = \frac{kx^2}{2} \qquad (5.19)\]
You can check this is right: taking \(-dV/dx\) gives \(-kx\), which is exactly the spring force. So the force is indeed the negative slope of the potential energy, as required.
Energy exchange in an oscillating spring. Now let the block of mass \(m\) be pulled out to \(x_m\) and released from rest. As it moves back and forth, mechanical energy is conserved (the surface is smooth). At any position \(x\) between \(-x_m\) and \(+x_m\), the total mechanical energy equals what it had at the turning point \(x_m\):
\[\frac{1}{2}kx_m^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2\]
This equation tells a story. At the turning point (\(x = x_m\)) the block is momentarily at rest, so all the energy is potential. As it moves toward the centre, potential energy converts into kinetic energy. The speed — and hence the kinetic energy — is largest at the equilibrium position, \(x = 0\), where the potential energy is zero:
\[\frac{1}{2}mv_m^2 = \frac{1}{2}kx_m^2\]
Solving for the maximum speed \(v_m\):
\[v_m = \sqrt{\frac{k}{m}}\,x_m\]
You can check the formula makes dimensional sense: \(k/m\) has dimensions of \([\text{T}^{-2}]\), so \(\sqrt{k/m}\,x_m\) has the dimensions of speed. Throughout the motion, kinetic energy converts into potential energy and back again, but the total mechanical energy stays constant. This is shown graphically in Fig. 5.8.
Figure to come
Fig. 5.8 – Two complementary parabolas versus x: potential energy V (lowest at centre, rising toward the ends) and kinetic energy K (highest at centre, falling to zero at the ends), with their sum E = K + V drawn as a constant horizontal line.
We should note that this is an idealised calculation. The spring was taken to be massless, and the surface was assumed frictionless.
A few remarks on conservative forces. Before refining the calculation, three points are worth making.
(i) Time is missing from these energy arguments. Energy methods told us how much the spring compresses, but not how long the compression takes. For timing information, we would need to solve Newton’s second law for the system directly.
(ii) Not all forces are conservative. Friction, for example, is a non-conservative force — the work it does depends on the path, and it cannot be stored and recovered. When friction is present, the simple conservation-of-mechanical-energy rule must be modified. This is illustrated in the next example.
(iii) The zero of potential energy is a matter of choice. We are free to decide where potential energy is counted as zero — for the spring we chose the equilibrium position, and for gravity near the ground we chose the Earth’s surface. The choice is made for convenience. But once the zero is fixed for a given problem, it must be used consistently throughout — you cannot switch reference points partway through a calculation.
Modifying conservation when a non-conservative force is present. Suppose the forces on a body are a conservative force \(F_c\) and a non-conservative force \(F_{nc}\). By the work-energy theorem, the total work equals the change in kinetic energy:
\[(F_c + F_{nc})\,\Delta x = \Delta K\]
But for the conservative part, the work equals the drop in potential energy: \(F_c\,\Delta x = -\Delta V\). Substituting this in and moving terms around,
\[\Delta(K + V) = F_{nc}\,\Delta x\]
\[\Delta E = F_{nc}\,\Delta x\]
where \(E = K + V\) is the total mechanical energy. Over a whole path, this becomes
\[E_f - E_i = W_{nc}\]
where \(W_{nc}\) is the total work done by the non-conservative forces along the path. So mechanical energy is no longer constant — it changes by exactly the work done by the non-conservative forces. And unlike the conservative case, \(W_{nc}\) depends on the particular path taken from the initial point \(i\) to the final point \(f\).
5.10 Power
So far we have asked how much work is done. Often it is just as important to ask how fast it is done. Two cranes may lift the same load to the same height — doing equal work — but the one that does it in half the time is clearly the more capable machine.
We use this idea in everyday judgements too. We call a person physically fit not just because they can climb four floors, but because they can climb them quickly. The quantity that captures this “how fast is work done” is called power.
Two engines can do the same total work, but the one that does it in less time is more powerful. So “power” in physics is not about how large a force is, but about how quickly work is delivered — force and speed together, as we will see.
Average power. The average power of a force is the total work \(W\) divided by the total time \(t\) taken to do it:
\[P_{av} = \frac{W}{t}\]
Instantaneous power. The power at a single instant — the instantaneous power — is found by shrinking the time interval toward zero. It is the limiting value of the average power, written as a derivative:
\[P = \frac{dW}{dt} \qquad (5.20)\]
We can rewrite instantaneous power in a very useful form. The small amount of work \(dW\) done by a force \(\vec{F}\) over a small displacement \(d\vec{r}\) is \(dW = \vec{F}\cdot d\vec{r}\). Dividing by \(dt\):
\[P = \vec{F}\cdot\frac{d\vec{r}}{dt}\]
But \(d\vec{r}/dt\) is just the velocity \(\vec{v}\), so
\[P = \vec{F}\cdot\vec{v} \qquad (5.21)\]
where \(\vec{v}\) is the instantaneous velocity of the body when the force acting is \(\vec{F}\). This compact form — power equals force dotted with velocity — is the one most used in problems.
A scalar quantity. Power, like work and energy, is a scalar. Its dimensions are \([\text{ML}^2\text{T}^{-3}]\). In SI units, power is measured in the watt (W), where one watt is one joule per second (\(1 \text{ W} = 1 \text{ J s}^{-1}\)).
There is indeed another unit of power still in common use, the horse-power (hp):
\[1 \text{ hp} = 746 \text{ W}\]
It is still used to describe the output of cars, motorbikes, and similar machines.
\[100 \text{ (watt)} \times 10 \text{ (hour)} = 1000 \text{ watt hour}\] \[= 1 \text{ kilowatt hour (kWh)}\] \[= 10^3 \text{ (W)} \times 3600 \text{ (s)}\] \[= 3.6 \times 10^6 \text{ J}\]
Our electricity bills measure energy consumed in units of kWh.
5.11 Collisions
In physics we study motion — how position changes with time. But we are just as interested in the quantities that do not change during a physical process. The conservation of momentum and of energy are two prime examples of such unchanging quantities. In this section we put both laws to work on a very common event: a collision.
Collisions are everywhere. Games such as billiards, marbles, and carrom are built entirely around them. To understand the physics, we study the collision of two masses in an idealised form.
Consider two masses \(m_1\) and \(m_2\). The particle \(m_1\) moves with speed \(v_{1i}\) (the subscript “\(i\)” standing for initial), and we take \(m_2\) to be at rest. Taking the second mass to be at rest costs us no generality — we can always choose our viewpoint so that one body is initially still. In this setup, the moving mass \(m_1\) strikes the stationary mass \(m_2\), as shown in Fig. 5.10.
Figure to come
Fig. 5.10 – Mass m1 approaching a stationary mass m2 with speed v1i; after collision, m1 moves off at angle θ1 with speed v1f and m2 moves off at angle θ2 with speed v2f.
After the collision, the two masses fly off in different directions. Our task is to find the relationships that connect the masses, the velocities, and the angles.
5.11.1 Elastic and Inelastic Collisions
Momentum is always conserved. In every collision, the total linear momentum of the system is conserved — the total momentum just before the collision equals the total momentum just after. Here is why.
When the two objects collide, they push on each other with mutual impulsive forces during the short collision time \(\Delta t\). These forces change each body’s momentum:
\[\Delta \vec{p}_1 = \vec{F}_{12}\,\Delta t\] \[\Delta \vec{p}_2 = \vec{F}_{21}\,\Delta t\]
Here \(\vec{F}_{12}\) is the force on the first particle from the second, and \(\vec{F}_{21}\) is the force on the second from the first. By Newton’s third law, these are equal and opposite: \(\vec{F}_{12} = -\vec{F}_{21}\). Therefore the two momentum changes cancel:
\[\Delta \vec{p}_1 + \Delta \vec{p}_2 = 0\]
This holds even though the forces vary in a complicated way during the collision. Because Newton’s third law is true at every instant, the total impulse on the first body is equal and opposite to that on the second, at all times. So the total momentum of the pair does not change.
Kinetic energy is not always conserved. The total kinetic energy of the system, unlike momentum, is not necessarily conserved in a collision. During impact, the bodies deform, and this can generate heat and sound. Part of the initial kinetic energy is then converted into these other forms of energy.
A helpful way to picture the deformation is to imagine a “compressed spring” between the two masses during contact, as suggested in Fig. 5.10B.
Figure to come
Fig. 5.10B – Two masses momentarily in contact with a compressed spring between them, representing the temporary deformation during a collision.
If this imagined spring springs back to its original shape without any energy loss, the total kinetic energy after the collision equals the total before — though during the collision itself the kinetic energy dips and recovers. Such a collision is called an elastic collision.
If instead the deformation is never relieved and the two bodies stick and move together afterward, the collision is called a completely inelastic collision.
The in-between case — where the deformation is only partly relieved and some kinetic energy is lost — is the most common of all, and is simply called an inelastic collision.
5.11.2 Collisions in One Dimension
Completely inelastic collision. Consider first a completely inelastic collision in one dimension. Then in Fig. 5.10 the angles vanish, \(\theta_1 = \theta_2 = 0\), and the two masses move together with a common final velocity \(v_f\). Momentum conservation gives
\[m_1 v_{1i} = (m_1 + m_2)v_f \qquad \text{(momentum conservation)}\]
\[v_f = \frac{m_1}{m_1 + m_2}v_{1i} \qquad (5.22)\]
The kinetic energy lost in this collision is the initial kinetic energy minus the final:
\[\Delta K = \frac{1}{2}m_1 v_{1i}^2 - \frac{1}{2}(m_1 + m_2)v_f^2\]
Substituting \(v_f\) from Eq. (5.22):
\[= \frac{1}{2}m_1 v_{1i}^2 - \frac{1}{2}\frac{m_1^2}{m_1 + m_2}v_{1i}^2 \qquad \text{[using Eq. (5.22)]}\]
\[= \frac{1}{2}m_1 v_{1i}^2\left[1 - \frac{m_1}{m_1 + m_2}\right]\]
\[= \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}v_{1i}^2\]
This is a positive quantity, as expected — kinetic energy is indeed lost when the bodies stick together.
Elastic collision. Now consider an elastic collision in one dimension, again with \(\theta_1 = \theta_2 = 0\). Here both momentum and kinetic energy are conserved, giving two equations:
\[m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f} \qquad (5.23)\]
\[m_1 v_{1i}^2 = m_1 v_{1f}^2 + m_2 v_{2f}^2 \qquad (5.24)\]
To solve these, rearrange each. From Eq. (5.23): \(m_1(v_{1i} - v_{1f}) = m_2 v_{2f}\). From Eq. (5.24): \(m_1(v_{1i}^2 - v_{1f}^2) = m_2 v_{2f}^2\). Dividing the second by the first, and using \(v_{1i}^2 - v_{1f}^2 = (v_{1i} - v_{1f})(v_{1i} + v_{1f})\), we get
\[v_{2f} = v_{1i} + v_{1f} \qquad (5.25)\]
Substituting this back into Eq. (5.23) and solving for the two final velocities:
\[v_{1f} = \frac{(m_1 - m_2)}{m_1 + m_2}v_{1i} \qquad (5.26)\]
\[v_{2f} = \frac{2m_1 v_{1i}}{m_1 + m_2} \qquad (5.27)\]
So the “unknowns” \(\{v_{1f}, v_{2f}\}\) are found entirely in terms of the “knowns” \(\{m_1, m_2, v_{1i}\}\). Two special cases are especially revealing.
Case I — equal masses (\(m_1 = m_2\)). Putting \(m_1 = m_2\) in Eqs. (5.26) and (5.27):
\[v_{1f} = 0\] \[v_{2f} = v_{1i}\]
The first mass stops dead, and the second mass moves off with exactly the incoming speed. The moving ball comes to rest and hands over its entire velocity to the target.
Case II — one mass dominates (\(m_2 \gg m_1\)). When the target is far heavier than the projectile:
\[v_{1f} \approx -v_{1i}\] \[v_{2f} \approx 0\]
The heavy mass barely moves, while the light mass bounces straight back with nearly its original speed reversed — much like a ball rebounding off a wall.
From one dimension to two. If the initial and final velocities of both bodies lie along the same straight line, the collision is called a one-dimensional collision, or a head-on collision. For small spherical bodies, this happens when the line of travel of body 1 passes through the centre of the stationary body 2. In general, though, a collision is two-dimensional, with the initial and final velocities lying in a plane.
5.11.3 Collisions in Two Dimensions
Figure 5.10 also represents the more general two-dimensional case: a moving mass \(m_1\) striking a stationary mass \(m_2\) and the two flying off at angles. Linear momentum is conserved here too. Since momentum is a vector, its conservation gives three equations, one for each direction \(\{x, y, z\}\).
We choose the \(x\)–\(y\) plane to be the plane containing the final velocity directions of \(m_1\) and \(m_2\). Conservation of the \(z\)-component of momentum then forces the entire collision to lie in this \(x\)–\(y\) plane. The \(x\)- and \(y\)-component equations of momentum conservation are
\[m_1 v_{1i} = m_1 v_{1f}\cos\theta_1 + m_2 v_{2f}\cos\theta_2 \qquad (5.28)\]
\[0 = m_1 v_{1f}\sin\theta_1 - m_2 v_{2f}\sin\theta_2 \qquad (5.29)\]
In most situations we know \(\{m_1, m_2, v_{1i}\}\). But there are four unknowns \(\{v_{1f}, v_{2f}, \theta_1, \theta_2\}\) and only these two equations. (If \(\theta_1 = \theta_2 = 0\), we recover the one-dimensional Eq. 5.23.)
If the collision is also elastic, kinetic energy conservation supplies one more equation:
\[\frac{1}{2}m_1 v_{1i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2 \qquad (5.30)\]
Even so, three equations cannot fix four unknowns — we are still one equation short. So at least one of the four unknowns, say \(\theta_1\), must be supplied from outside for the problem to be solvable. In an experiment, \(\theta_1\) could be measured by sweeping a detector in an arc from the \(x\)-axis toward the \(y\)-axis. Once \(\{m_1, m_2, v_{1i}, \theta_1\}\) are known, Eqs. (5.28)–(5.30) determine the remaining \(\{v_{1f}, v_{2f}, \theta_2\}\).
When do bodies actually “collide”? The analysis simplifies greatly if we treat the bodies as smooth spheres that interact only at the instant they touch — which is a good picture for marbles, carrom coins, and billiard balls.
But collisions need not involve contact. Consider a comet swinging in from deep space toward the Sun, or an alpha particle approaching an atomic nucleus and veering away. Here the bodies interact through forces that act at a distance, without ever touching. Such an event is called scattering. The speeds and directions in which the two particles finally move apart depend on their initial velocities, on the nature of the force between them, and on their masses, shapes, and sizes.
5.12 Summary
- The work-energy theorem states that the change in kinetic energy of a body equals the work done by the net force on the body.
\[K_f - K_i = W_{net}\]
A force is conservative if either (i) the work done by it on an object is path-independent and depends only on the end points \(\{x_i, x_f\}\), or (ii) the work done by the force is zero over any arbitrary closed path — that is, when the object returns to its initial position.
For a conservative force in one dimension, we can define a potential energy function \(V(x)\) such that
\[F(x) = -\frac{dV(x)}{dx}\]
\[\text{or} \qquad V_i - V_f = \int_{x_i}^{x_f} F(x)\,dx\]
The principle of conservation of mechanical energy states that the total mechanical energy of a body remains constant if the only forces acting on the body are conservative.
The gravitational potential energy of a particle of mass \(m\) at a height \(x\) above the Earth’s surface is
\[V(x) = mgx\]
where the variation of \(g\) with height is ignored.
- The elastic potential energy of a spring of force constant \(k\) and extension \(x\) is
\[V(x) = \frac{1}{2}kx^2\]
- The scalar (dot) product of two vectors \(\vec{A}\) and \(\vec{B}\) is written \(\vec{A}\cdot\vec{B}\) and is a scalar, given by
\[\vec{A}\cdot\vec{B} = AB\cos\theta\]
where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\). It can be positive, negative, or zero depending on the value of \(\theta\). The scalar product can be read as the magnitude of one vector times the component of the other vector along the first. For the unit vectors:
\[\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1 \quad \text{and} \quad \hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0\]
Scalar products obey the commutative and the distributive laws.
Table of Physical Quantities
| Physical Quantity | Symbol | Dimensions | Units | Remarks |
|---|---|---|---|---|
| Work | \(W\) | \([\text{ML}^2\text{T}^{-2}]\) | J | \(W = \vec{F}\cdot\vec{d}\) |
| Kinetic energy | \(K\) | \([\text{ML}^2\text{T}^{-2}]\) | J | \(K = \dfrac{1}{2}mv^2\) |
| Potential energy | \(V(x)\) | \([\text{ML}^2\text{T}^{-2}]\) | J | \(F(x) = -\dfrac{dV(x)}{dx}\) |
| Mechanical energy | \(E\) | \([\text{ML}^2\text{T}^{-2}]\) | J | \(E = K + V\) |
| Spring constant | \(k\) | \([\text{MT}^{-2}]\) | \(\text{N m}^{-1}\) | \(F = -kx\,;\ \ V(x) = \dfrac{1}{2}kx^2\) |
| Power | \(P\) | \([\text{ML}^2\text{T}^{-3}]\) | W | \(P = \vec{F}\cdot\vec{v}\,;\ \ P = \dfrac{dW}{dt}\) |
5.13 Points to Ponder
The phrase “calculate the work done” is, by itself, incomplete. We must always refer (or make clear from the context) to the work done by a specific force — or a specific group of forces — on a given body, over a certain displacement.
Work done is a scalar quantity, but unlike mass and kinetic energy (which are always positive scalars), work can be positive or negative. The work done by friction or a viscous force on a moving body is negative.
For two bodies, the sum of the mutual forces they exert on each other is zero, by Newton’s third law:
\[\vec{F}_{12} + \vec{F}_{21} = 0\]
However, the sum of the work done by these two forces need not cancel:
\[W_{12} + W_{21} \neq 0\]
(though it may sometimes turn out to be zero).
The work done by a force can sometimes be found even when the exact nature of the force is unknown. This is clear from Example 5.2, where the work-energy theorem was used in exactly such a situation (the unknown resistive force on a raindrop).
The work-energy theorem is not independent of Newton’s second law — it may be viewed as a scalar form of that law. Likewise, the principle of conservation of mechanical energy may be viewed as a consequence of the work-energy theorem for conservative forces.
The work-energy theorem holds in all inertial frames. It can also be extended to non-inertial frames, provided we include the pseudoforces when calculating the net force on the body.
The potential energy of a body under a conservative force is always undetermined up to a constant — the point where potential energy is taken as zero is a matter of choice. For the gravitational potential energy \(mgh\), the zero is usually chosen at the ground. For the spring potential energy \(kx^2/2\), the zero is chosen at the equilibrium position of the oscillating mass.
Not every force in mechanics has an associated potential energy. For example, the work done by friction over a closed path is not zero, so no potential energy can be defined for friction.
During a collision: (a) the total linear momentum is conserved at every instant of the collision; (b) kinetic energy conservation (even for an elastic collision) applies only after the collision is over — it does not hold at every instant during the collision. In fact, the two colliding objects are deformed during contact and may even be momentarily at rest relative to each other.
5.14 NCERT Questions
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
- work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
- work done by gravitational force in the above case.
- work done by friction on a body sliding down an inclined plane.
- work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
- work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the
- work done by the applied force in 10 s,
- work done by friction in 10 s,
- work done by the net force on the body in 10 s,
- change in kinetic energy of the body in 10 s, and interpret your results.
Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
The potential energy function for a particle executing linear simple harmonic motion is given by \(V(x) = kx^2/2\), where \(k\) is the force constant of the oscillator. For \(k = 0.5 \text{ N m}^{-1}\), the graph of \(V(x)\) versus \(x\) is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this potential must “turn back” when it reaches \(x = \pm 2\) m.
Answer the following:
- The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
- Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
- An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
- In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?
- Underline the correct alternative:
- When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
- Work done by a body against friction always results in a loss of its kinetic/potential energy.
- The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
- In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
- State if each of the following statements is true or false. Give reasons for your answer.
- In an elastic collision of two bodies, the momentum and energy of each body is conserved.
- Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
- Work done in the motion of a body over a closed loop is zero for every force in nature.
- In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
- Answer carefully, with reasons:
- In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
- Is the total linear momentum conserved during the short time of an elastic collision of two balls?
- What are the answers to (a) and (b) for an inelastic collision?
- If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy.)
- A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time \(t\) is proportional to
- \(t^{1/2}\) (ii) \(t\) (iii) \(t^{3/2}\) (iv) \(t^{2}\)
- A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time \(t\) is proportional to
- \(t^{1/2}\) (ii) \(t\) (iii) \(t^{3/2}\) (iv) \(t^{2}\)
A body constrained to move along the \(z\)-axis of a coordinate system is subject to a constant force \(\vec{F}\) given by \[\vec{F} = -\hat{i} + 2\hat{j} + 3\hat{k} \ \text{N}\] where \(\hat{i}, \hat{j}, \hat{k}\) are unit vectors along the \(x\)-, \(y\)- and \(z\)-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the \(z\)-axis?
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass \(= 9.11\times10^{-31}\) kg, proton mass \(= 1.67\times10^{-27}\) kg, \(1 \text{ eV} = 1.60\times10^{-19}\) J).
A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is \(10 \text{ m s}^{-1}\)?
A molecule in a gas container hits a horizontal wall with speed \(200 \text{ m s}^{-1}\) and angle \(30^\circ\) with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
A pump on the ground floor of a building can pump up water to fill a tank of volume \(30 \text{ m}^3\) in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed \(V\). If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision?
The bob A of a pendulum released from \(30^\circ\) to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of \(0.05 \text{ kg s}^{-1}\). What is the speed of the trolley after the entire sand bag is empty?
A body of mass 0.5 kg travels in a straight line with velocity \(v = a x^{3/2}\) where \(a = 5 \text{ m}^{-1/2}\text{ s}^{-1}\). What is the work done by the net force during its displacement from \(x = 0\) to \(x = 2\) m?
The blades of a windmill sweep out a circle of area \(A\). (a) If the wind flows at a velocity \(v\) perpendicular to the circle, what is the mass of the air passing through it in time \(t\)? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that \(A = 30 \text{ m}^2\), \(v = 36\) km/h and the density of air is \(1.2 \text{ kg m}^{-3}\). What is the electrical power produced?
A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force? (b) Fat supplies \(3.8 \times 10^7\) J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house.
5.15 Check Your Concepts
The work done by a force can be zero even when both the force and the displacement are nonzero. Using the scalar product \(\vec{F}\cdot\vec{d} = Fd\cos\theta\), explain under what condition this happens, and give one physical example.
A ball is thrown vertically upward and returns to the thrower’s hand. Ignoring air resistance, use the work-energy theorem to explain why the ball’s speed at any given height on the way up is the same as its speed at that height on the way down.
A block is lowered slowly at constant velocity by a rope. State the sign of the work done by (a) the gravitational force and (b) the tension in the rope, and explain your reasoning for each.
Two cars of equal mass travel such that one moves at twice the speed of the other. Compare (a) their kinetic energies and (b) the distances they take to stop under equal braking forces. Justify your answer using the work-energy theorem.
State the three equivalent tests used to decide whether a force is conservative. Which of these tests most easily shows that friction is non-conservative, and why?
Explain why the point at which potential energy is taken to be zero can be chosen freely, and why, once chosen, it must not be changed partway through a calculation.
In a completely inelastic collision between two bodies, is linear momentum conserved? Is kinetic energy conserved? Give a clear reason for each answer.
“A force that is always perpendicular to a body’s velocity can change the body’s direction but never its speed.” State whether this is true, and explain it using the concept of work done.
A 60 W bulb and a 100 W bulb are switched on for the same length of time. Which one consumes more electrical energy, and why? In your answer, clearly distinguish between power and energy.
Using the results for a one-dimensional elastic collision between equal masses, explain why, in a Newton’s cradle, releasing a single ball at one end causes exactly one ball to swing out at the other end.
5.16 Practice with Numericals
Two vectors are given by \(\vec{A} = 2\hat{i} + 3\hat{j} + \hat{k}\) and \(\vec{B} = \hat{i} - \hat{j} + \hat{k}\). Find (a) the scalar product \(\vec{A}\cdot\vec{B}\), (b) the angle between them, and (c) the projection of \(\vec{A}\) on \(\vec{B}\).
A variable force acting along the \(x\)-axis is given by \(F = (2x + 3)\) N, where \(x\) is in metres. Find the work done by this force in moving a body from \(x = 1\) m to \(x = 4\) m.
A 0.20 kg ball moving at \(10 \text{ m s}^{-1}\) is brought to rest over a distance of 2.0 m by a constant resisting force. Using the work-energy theorem, find the magnitude of the resisting force.
A 1.0 kg stone is dropped from rest at the top of a 20 m high cliff (take \(g = 10 \text{ m s}^{-2}\)). Using conservation of mechanical energy, find its speed when it is 5.0 m above the ground.
A spring of force constant \(500 \text{ N m}^{-1}\) is compressed by 0.20 m and a 2.0 kg block is held against it on a frictionless surface. Find (a) the elastic potential energy stored in the spring, and (b) the speed the block acquires when released from rest.
A pump lifts 600 kg of water to a height of 15 m in 20 s (take \(g = 10 \text{ m s}^{-2}\)). Find the useful power output of the pump.
A 3.0 kg body moving at \(4.0 \text{ m s}^{-1}\) collides head-on with a stationary 1.0 kg body and the two stick together. Find (a) their common velocity after the collision and (b) the fraction of the initial kinetic energy lost.
A 2.0 kg ball moving at \(6.0 \text{ m s}^{-1}\) makes a head-on elastic collision with a stationary 4.0 kg ball. Find the velocity of each ball after the collision.
A car with an engine of output power 40 kW moves along a level road against a constant resistive force of 800 N. Find the maximum speed the car can reach on this road.
A 0.050 kg bullet moving at \(400 \text{ m s}^{-1}\) strikes and embeds itself in a 2.0 kg wooden block resting on a frictionless surface. Find (a) the speed of the block-and-bullet just after impact and (b) the kinetic energy lost in the process.