NCERT Solutions Class 11 Physics Chapter 5 — Work, Energy and Power

NCERT Class 11 Physics — Work, Energy and Power, Chapter 5 Exercises. All 23 questions solved.

The exercises at the end of Chapter 5 fall into four groups. Questions 5.1 to 5.8 test ideas: the sign of work, potential energy curves, conservative forces and what is conserved in a collision. Questions 5.9 to 5.11, 5.15 and 5.21 to 5.23 are about work and power; 5.12, 5.13, 5.18 and 5.20 use the work-energy theorem; and 5.14, 5.16, 5.17 and 5.19 are about collisions and momentum. Almost all the arithmetic runs on three results:

• Work done by a constant force — $W = \vec F\cdot\vec d = Fd\cos\theta$. It is positive when the force has a component along the displacement and negative when the component is against it.
• The work-energy theorem — the work done by the net force equals the change in kinetic energy, $W_{\text{net}} = \Delta K$ with $K = \tfrac12mv^2$. When only conservative forces act, the mechanical energy $K + V$ stays constant.
• Power — $P = \dfrac{dW}{dt} = \vec F\cdot\vec v$, and the efficiency of a machine is its useful output divided by its input.

We take $g = 9.8\ \text{m s}^{-2}$ throughout.

Key insight. Compare the start and the end before you look at the forces in between. In this exercise that settles most answers: the work-energy theorem gives the work done by air resistance on a raindrop without knowing the resistive force (5.13), the speed of a pendulum bob without following its path (5.18) and the work of a varying force from two speeds (5.20), while momentum and kinetic energy together leave only one possible outcome of a collision (5.16, 5.17). Ask what is conserved first, then whether any force takes energy out of the system.

Question 5.1

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: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Solution. In every part the sign is the sign of $\cos\theta$, where $\theta$ is the angle between the force and the displacement.

(a) The man pulls the rope upwards and the bucket moves upwards, so $\theta = 0$ and the work is positive.

(b) Gravity pulls the bucket down while it moves up, so $\theta = 180^\circ$ and the work is negative.

(c) Friction always opposes the sliding. The body moves down the incline and friction acts up it, so $\theta = 180^\circ$ and the work is negative.

(d) To keep the body moving at constant velocity, the applied force must balance friction, so it points along the motion: the work is positive. The net work is zero, because friction does an equal amount of negative work, which is why the kinetic energy does not change.

(e) Air resistance always acts against the bob’s velocity, whichever way it swings, so it does negative work on every swing. That negative work is what removes the pendulum’s energy and brings it to rest.

Question 5.2

A body of mass $2\ \text{kg}$ initially at rest moves under the action of an applied horizontal force of $7\ \text{N}$ on a table with coefficient of kinetic friction $= 0.1$. Compute the (a) work done by the applied force in $10\ \text{s}$, (b) work done by friction in $10\ \text{s}$, (c) work done by the net force on the body in $10\ \text{s}$, (d) change in kinetic energy of the body in $10\ \text{s}$, and interpret your results.

Solution. Every part needs the distance travelled in $10\ \text{s}$, so find the acceleration first. The table is horizontal, so the normal reaction equals the weight and the kinetic friction is

$$f = \mu mg = 0.1\times2\times9.8 = 1.96\ \text{N}$$

The net force is $7 – 1.96 = 5.04\ \text{N}$, so $a = 5.04/2 = 2.52\ \text{m s}^{-2}$. Starting from rest, in $10\ \text{s}$ the body travels

$$s = \tfrac12at^2 = \tfrac12\times2.52\times10^2 = 126\ \text{m}$$

and reaches $v = at = 25.2\ \text{m s}^{-1}$.

(a) The applied force is along the motion: $W_F = 7\times126 = 882\ \text{J}$.

(b) Friction is against the motion: $W_f = -1.96\times126 = -246.96 \approx -247\ \text{J}$.

(c) The net force is along the motion: $W_{\text{net}} = 5.04\times126 = 635\ \text{J}$, which is also $882 – 247$.

(d) $\Delta K = \tfrac12mv^2 – 0 = \tfrac12\times2\times(25.2)^2 = 635\ \text{J}$.

Interpretation. The work done by the net force, $635\ \text{J}$, is exactly the change in kinetic energy. That is the work-energy theorem. Of the $882\ \text{J}$ supplied by the applied force, $247\ \text{J}$ is lost against friction as heat and the remaining $635\ \text{J}$ becomes kinetic energy.

Question 5.3

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.

Solution. The total energy is $E = K + V(x)$, and kinetic energy can never be negative. So the particle can only be where $V(x) \le E$; wherever $V(x) > E$ it would need negative kinetic energy, so it cannot be there. For the same reason, the least total energy it can have is the lowest value $V(x)$ takes anywhere. The four graphs of Fig. 5.11 are taken in order from the top as (a) to (d), which is how the key labels them. In each graph below, the red band on the $x$-axis marks the region the particle cannot enter.

Question 5.3 (a)

Solution. For $x < a$ the potential energy is zero, below $E$, so the particle can move there. For $x > a$ it is $V_0$, and the cross shows $E < V_0$, so the particle cannot be found in $x > a$: it reaches the step and turns back. The lowest value of $V(x)$ is $0$, so the total energy must be at least $0$. A ball rolling along a floor towards a raised platform that it has too little energy to climb onto is a physical example.

Question 5.3 (b)

Solution. Every level of this staircase, including the lowest one to the left of the origin, lies above the cross that marks $E$. So $V(x) > E$ for every $x$, and a particle with this energy cannot be found anywhere. To exist at all, it needs a total energy at least equal to the lowest level of $V(x)$, the one for $x < 0$.

Note on the book’s answer. The key gives the minimum energy as $V_1$. The figure never labels a $V_1$; the key’s $V_1$ is the lowest level of this graph, the segment to the left of the origin.

Question 5.3 (c)

Solution. Outside the interval from $a$ to $b$ the potential energy is $V_0$, which is more than $E$, so the particle cannot be there. Between $a$ and $b$ it is $-V_1$, below $E$, so the particle moves freely inside that interval, bouncing between the two walls. The lowest value of $V(x)$ is $-V_1$, so that is the minimum total energy. This is a square potential well; an electron trapped in a small region, or a ball in a flat-bottomed pit with vertical sides, behaves this way.

Question 5.3 (d)

Solution. The dotted line at the level of $E$ meets the falling curves at $x = \pm b/2$. Between $a/2$ and $b/2$ on the right, and between $-b/2$ and $-a/2$ on the left, the curve is above $E$, so the particle cannot be there. It can be inside the well, $|x| < a/2$, where $V = -V_1 < E$, or outside, $|x| > b/2$, where the curve has fallen below $E$. The lowest value of $V(x)$ is $-V_1$, so that is the minimum total energy. This is the shape of the potential energy of an $\alpha$-particle in a nucleus: a deep well held in by a barrier. Classically, a particle inside with energy $E$ could never cross the barrier; that $\alpha$-particles escape anyway is one of the effects explained by quantum mechanics.

Question 5.4

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 $1\ \text{J}$ moving under this potential must ‘turn back’ when it reaches $x = \pm2\ \text{m}$.

Solution. The total energy $E = K + V(x)$ stays fixed at $1\ \text{J}$, and $K = \tfrac12mv^2$ can never be negative, so the particle can only be where $V(x) \le 1\ \text{J}$. The boundary is where all the energy is potential:

$$\tfrac12kx^2 = E \quad\Rightarrow\quad x^2 = \frac{2E}{k} = \frac{2\times1}{0.5} = 4\ \text{m}^2 \quad\Rightarrow\quad x = \pm2\ \text{m}$$

At $x = \pm2\ \text{m}$ the kinetic energy is zero, so the particle is momentarily at rest. Going any further would make $V(x) > 1\ \text{J}$ and $K$ negative, which is impossible. The restoring force $F = -kx$ points back towards the origin there, so the particle turns back. Hence it oscillates between $x = -2\ \text{m}$ and $x = +2\ \text{m}$.

Question 5.5

Answer the following: (a) 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? (b) 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? (c) 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? (d) In Fig. 5.13(i) the man walks $2\ \text{m}$ carrying a mass of $15\ \text{kg}$ 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 $15\ \text{kg}$ hangs at its other end. In which case is the work done greater?

Question 5.5 (a)

Solution. Friction from the air acts on the rocket against its motion, so it does negative work on the rocket and takes away some of the rocket’s mechanical energy. That energy appears as the heat that burns the casing. So the heat comes from the rocket’s own energy, its kinetic and potential energy, and so ultimately from its fuel, not from the atmosphere.

Question 5.5 (b)

Solution. Gravity is a conservative force, so the work it does depends only on the starting and finishing points: $W = -\Delta V$, where $V$ is the gravitational potential energy. Over one complete orbit the comet comes back to the same point, so $\Delta V = 0$ and the total work is zero. On the way out from the sun gravity does negative work, slowing the comet; on the way back it does an equal amount of positive work.

Question 5.5 (c)

Solution. Air resistance removes energy, so the total energy $K + V$ falls and the satellite drops into a lower orbit. As it comes closer to the earth, its gravitational potential energy falls, and it falls by more than the energy lost to friction. The difference goes into kinetic energy, so the satellite speeds up. For a circular orbit of radius $r$ (Chapter 7), $K = GMm/2r$ and $V = -GMm/r$: when $r$ decreases, $V$ drops by twice as much as $K$ rises, and half of that drop is dissipated as heat.

Question 5.5 (d)

Solution. In case (i) the man’s hands push the mass straight up to hold it against gravity, while the mass moves horizontally. The force is perpendicular to the displacement, so the work done on the mass is $mgd\cos90^\circ = 0$.

In case (ii) the rope is horizontal at the man’s end, and to raise the hanging mass steadily he must pull with a force equal to its weight, along the direction he walks. The work is $Fd = mgd = 15\times9.8\times2 = 294\ \text{J}$, which is exactly the potential energy gained by the mass as it rises $2\ \text{m}$.

Question 5.6

Underline the correct alternative: (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered. (b) Work done by a body against friction always results in a loss of its kinetic/potential energy. (c) 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. (d) 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.

Solution. (a) For a conservative force $W = -\Delta V$, so positive work means $\Delta V < 0$: the potential energy decreases, as when gravity does positive work on a falling body.

(b) Friction opposes motion and does negative work on the body, so by the work-energy theorem the body loses kinetic energy.

(c) The internal forces come in equal and opposite pairs by Newton’s third law and cancel in the total, so only the external force changes the total momentum: $d\vec P/dt = \vec F_{\text{ext}}$.

(d) In any collision the forces between the two bodies are internal, so the total linear momentum is unchanged. Kinetic energy is not conserved in an inelastic collision, since part of it becomes heat or deformation. The total energy, counting all forms, is also conserved provided the two bodies form an isolated system.

Question 5.7

State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Question 5.7 (a)

Solution. A collision exists precisely to exchange momentum and energy between the bodies: each body pushes the other, so each one’s momentum and kinetic energy change. What is conserved is the total momentum and, in an elastic collision, the total kinetic energy of the pair.

Question 5.7 (b)

Solution. External forces can do work on the system and so add energy to it or take energy away. A body pushed along the floor gains energy from the push. The total energy is conserved only for an isolated system, one on which no external force does work.

Question 5.7 (c)

Solution. The work over a closed loop is zero only for conservative forces, such as gravity or the force of a spring. For a non-conservative force such as friction it is not: friction opposes the motion on every part of the loop, so its work is negative all the way round and the total is negative.

Question 5.7 (d)

Solution. In most inelastic collisions some kinetic energy becomes heat, sound or deformation, and the final kinetic energy is less. But that is not guaranteed. If the collision releases energy stored inside the bodies — for instance, if the impact sets off an explosive charge, or releases a compressed spring, or triggers a nuclear reaction — the final kinetic energy can be more than the initial. “Inelastic” only means that kinetic energy is not conserved.

Question 5.8

Answer carefully, with reasons: (a) 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)? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls? (c) What are the answers to (a) and (b) for an inelastic collision? (d) 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).

Question 5.8 (a)

Solution. While the balls are in contact they are squashed slightly, and part of their kinetic energy is stored as elastic potential energy of deformation. So during the contact the kinetic energy is less than before. In an elastic collision all of it is returned as the balls spring apart, which is why the kinetic energy after the collision equals that before.

Question 5.8 (b)

Solution. The forces the balls exert on each other are internal, equal and opposite at every instant, so they cannot change the total momentum at any moment, during the contact or after it.

Question 5.8 (c)

Solution. The momentum argument in (b) never used the kind of collision, so the total momentum is conserved during and after an inelastic collision too. Kinetic energy is not conserved during the contact, as in (a), and in an inelastic collision it is not restored afterwards either, because some of it has become heat or permanent deformation.

Question 5.8 (d)

Solution. If the potential energy depends only on the separation, the force between the balls is conservative. As they approach, kinetic energy becomes potential energy; as they separate back to the same distance, the potential energy returns to its earlier value, so all the kinetic energy is recovered. Nothing is lost, so the collision is elastic.

Question 5.9

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

Solution. Constant acceleration means a constant force $F = ma$. Starting from rest, the velocity grows as $v = at$. The power delivered is force times velocity:

$$P = Fv = (ma)(at) = ma^2t$$

With $m$ and $a$ constant, $P \propto t$.

Note on the book’s answer. The key labels this answer (b); it is the second option, (ii) $t$.

Question 5.10

A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time $t$ is proportional to

Solution. With constant power the force is not constant, so the constant-acceleration equations do not apply. Use energy instead: a constant power $P$ delivers work $Pt$ in time $t$, and if the body starts from rest, all of it becomes kinetic energy:

$$\tfrac12mv^2 = Pt \quad\Rightarrow\quad v = \sqrt{\frac{2P}{m}}\;t^{1/2}$$

The displacement is the integral of the velocity:

$$x = \int_0^t \sqrt{\frac{2P}{m}}\;t^{1/2}\,dt = \sqrt{\frac{2P}{m}}\cdot\frac23\,t^{3/2} \quad\Rightarrow\quad x \propto t^{3/2}$$

Note on the book’s answer. The key labels this answer (c); it is the third option, (iii) $t^{3/2}$.

Question 5.11

A body constrained to move along the $z$-axis of a coordinate system is subject to a constant force $\mathbf F$ given by

$$\mathbf F = -\hat{\imath} + 2\hat{\jmath} + 3\hat k\ \text{N}$$

where $\hat{\imath}, \hat{\jmath}, \hat k$ 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 $4\ \text{m}$ along the $z$-axis?

Solution. Work is the dot product of force and displacement, so only the component of the force along the displacement counts. The displacement is $\vec d = 4\hat k\ \text{m}$; the $x$- and $y$-components of the force are perpendicular to it and do no work (they are balanced by whatever constrains the body to the $z$-axis):

$$W = \mathbf F\cdot\vec d = (-\hat{\imath} + 2\hat{\jmath} + 3\hat k)\cdot(4\hat k) = 3\times4 = 12\ \text{J}$$

Question 5.12

An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy $10\ \text{keV}$, and the second with $100\ \text{keV}$. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass $= 9.11\times10^{-31}\ \text{kg}$, proton mass $= 1.67\times10^{-27}\ \text{kg}$, $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$).

Solution. From $K = \tfrac12mv^2$, the speed is $v = \sqrt{2K/m}$. The proton has ten times the kinetic energy, but it is about 1800 times as heavy, and mass matters more. Taking the ratio, the factor 2 cancels:

$$\frac{v_e}{v_p} = \sqrt{\frac{K_e}{K_p}\cdot\frac{m_p}{m_e}} = \sqrt{\frac{10}{100}\times\frac{1.67\times10^{-27}}{9.11\times10^{-31}}} = \sqrt{0.1\times1833} = \sqrt{183.3} = 13.5$$

For the actual speeds: $v_e = \sqrt{2\times10^4\times1.60\times10^{-19}/9.11\times10^{-31}} = 5.93\times10^7\ \text{m s}^{-1}$ and $v_p = \sqrt{2\times10^5\times1.60\times10^{-19}/1.67\times10^{-27}} = 4.38\times10^6\ \text{m s}^{-1}$, whose ratio is again $13.5$.

Question 5.13

A rain drop of radius $2\ \text{mm}$ falls from a height of $500\ \text{m}$ 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 $10\ \text{m s}^{-1}$?

Solution. First the mass of the drop, taking the density of water as $1000\ \text{kg m}^{-3}$:

$$m = \tfrac43\pi r^3\rho = \tfrac43\pi(2\times10^{-3})^3\times1000 = 3.35\times10^{-5}\ \text{kg}$$

Gravity. The weight is constant, so it does the same work on each $250\ \text{m}$ half, however the speed changes:

$$W_g = mgh = 3.35\times10^{-5}\times9.8\times250 = 0.082\ \text{J}$$

Resistive force. This force changes as the drop speeds up, so its work cannot be found from force times distance. The work-energy theorem gets round that: the total work of all the forces equals the change in kinetic energy. The drop starts at rest and lands with

$$K = \tfrac12mv^2 = \tfrac12\times3.35\times10^{-5}\times10^2 = 1.68\times10^{-3}\ \text{J}$$

so

$$W_r = K – W_g(\text{total}) = 1.68\times10^{-3} – 2\times0.0821 = -0.1625\ \text{J}$$

It is negative because the resistance opposes the fall. Almost all the energy gravity supplies is dissipated; only about 1% reaches the ground as kinetic energy. The same idea splits the total. In the second half the speed is constant, so the net work there is zero and the resistance does $-0.082\ \text{J}$, cancelling gravity; the rest, about $-0.080\ \text{J}$, is done in the first half.

Question 5.14

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?

Solution. Resolve the molecule’s velocity along the wall and along the normal. The component along the wall, $200\sin30^\circ = 100\ \text{m s}^{-1}$, is unchanged. The normal component, $200\cos30^\circ = 173\ \text{m s}^{-1}$, is reversed. So the molecule’s own momentum changes by $2mv\cos30^\circ$, directed away from the wall: the momentum of the molecule alone is not conserved.

The momentum of the molecule and the wall together is conserved, because the forces between them are internal. The wall receives an equal and opposite momentum, $2mv\cos30^\circ$ into the wall. Its mass is so enormous compared with the molecule’s that this gives it a negligible velocity.

The molecule rebounds with the same speed, so its kinetic energy is unchanged, and the wall’s recoil energy is negligible. Kinetic energy is conserved, so the collision is elastic.

Question 5.15

A pump on the ground floor of a building can pump up water to fill a tank of volume $30\ \text{m}^3$ in $15\ \text{min}$. If the tank is $40\ \text{m}$ above the ground, and the efficiency of the pump is $30\%$, how much electric power is consumed by the pump?

Solution. The useful work is the potential energy given to the water. Its mass is $30\ \text{m}^3\times1000\ \text{kg m}^{-3} = 3.0\times10^4\ \text{kg}$, so

$$W = mgh = 3.0\times10^4\times9.8\times40 = 1.176\times10^7\ \text{J}$$

This is done in $15\ \text{min} = 900\ \text{s}$, so the useful output power is

$$P_{\text{out}} = \frac{1.176\times10^7}{900} = 1.307\times10^4\ \text{W}$$

Only $30\%$ of the electrical input becomes useful work, so the input is larger than the output:

$$P_{\text{in}} = \frac{P_{\text{out}}}{0.30} = 4.36\times10^4\ \text{W}$$

Question 5.16

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?

Solution. An elastic collision must conserve both the total momentum and the total kinetic energy. Before the collision, with each ball of mass $m$, the momentum is $mV$ and the kinetic energy is $\tfrac12mV^2$. Test each outcome against both.

(i) Momentum: $2m\times\dfrac V2 = mV$, which is correct. Kinetic energy: $\tfrac12(2m)\left(\dfrac V2\right)^2 = \tfrac14mV^2$, only half of the original, so this is ruled out.

(ii) Momentum: $mV$. Kinetic energy: $\tfrac12mV^2$. Both are conserved.

(iii) Momentum: $3m\times\dfrac V3 = mV$, which is correct. Kinetic energy: $\tfrac12(3m)\left(\dfrac V3\right)^2 = \tfrac16mV^2$, so this is ruled out; it is what would happen if the balls stuck together, a perfectly inelastic collision.

Only (ii) conserves both. Physically, ball 1 hands its whole velocity to ball 2, which hands it on to ball 3, as equal masses do in an elastic head-on collision. This is the familiar Newton’s cradle.

Note on the book’s answer. The key gives (b), meaning the second outcome, (ii).

Question 5.17

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.

Solution. Bob A swings down and reaches the bottom of its arc, moving horizontally, just as it meets B; the collision is head-on. In an elastic head-on collision between equal masses, one of which is at rest, the two simply exchange velocities. Conservation of momentum and of kinetic energy give

$$mu = mv_A + mv_B, \qquad \tfrac12mu^2 = \tfrac12mv_A^2 + \tfrac12mv_B^2$$

whose solution (other than no collision at all) is $v_A = 0$, $v_B = u$. So A stops dead at the lowest point and B moves off along the table with A’s speed. With no speed left, A does not rise at all. The length of the pendulum was never given because it is not needed.

Question 5.18

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is $1.5\ \text{m}$, 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?

Solution. Released from the horizontal, the bob falls through a height equal to the length of the string, $h = 1.5\ \text{m}$. Measuring potential energy from the lowest point, its initial energy is all potential, $mgh$. Of this, $5\%$ is dissipated, so $95\%$ becomes kinetic energy at the bottom:

$$\tfrac12mv^2 = 0.95\,mgh \quad\Rightarrow\quad v = \sqrt{2\times0.95\times9.8\times1.5} = \sqrt{27.9} = 5.3\ \text{m s}^{-1}$$

The mass cancels, which is why it was not given.

Question 5.19

A trolley of mass $300\ \text{kg}$ carrying a sandbag of $25\ \text{kg}$ is moving uniformly with a speed of $27\ \text{km/h}$ 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?

Solution. The track is frictionless, and nothing pushes the trolley horizontally. The sand falls out through the floor; as it leaves, it is moving forwards at the trolley’s speed, and it carries its own momentum away with it. It exerts no horizontal force on the trolley, so the trolley’s velocity does not change. (Contrast a rocket, whose exhaust is pushed out backwards at high speed and so pushes the rocket forwards.) The leak rate only tells us how long the bag takes to empty, $25/0.05 = 500\ \text{s}$; it does not affect the speed.

Question 5.20

A body of mass $0.5\ \text{kg}$ 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 $x = 2\ \text{m}$?

Solution. The net force varies with $x$, but the work-energy theorem means we never need to find it: the work done by the net force is the change in kinetic energy, and the velocity is given at every $x$.

At $x = 0$: $v = 0$. At $x = 2\ \text{m}$: $v = 5\times2^{3/2} = 10\sqrt2 \approx 14.1\ \text{m s}^{-1}$, so $v^2 = 200\ \text{m}^2\,\text{s}^{-2}$.

$$W = \tfrac12mv^2 – 0 = \tfrac12\times0.5\times200 = 50\ \text{J}$$

Question 5.21

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 $A = 30\ \text{m}^2$, $v = 36\ \text{km/h}$ and the density of air is $1.2\ \text{kg m}^{-3}$. What is the electrical power produced?

Question 5.21 (a)

Solution. In time $t$ the air moves a distance $vt$, so the air that passes through the circle fills a cylinder of cross-section $A$ and length $vt$. Its volume is $Avt$ and its mass is density times volume.

Question 5.21 (b)

Solution. All of this air moves at speed $v$:

$$K = \tfrac12mv^2 = \tfrac12(\rho Avt)v^2 = \tfrac12\rho Av^3t$$

The energy goes as $v^3$: doubling the wind speed puts twice as much air through, each kilogram carrying four times the energy.

Question 5.21 (c)

Solution. Dividing (b) by $t$ gives the power carried by the wind, $\tfrac12\rho Av^3$. Convert $v = 36\ \text{km/h} = 10\ \text{m s}^{-1}$:

$$P_{\text{wind}} = \tfrac12\times1.2\times30\times10^3 = 1.8\times10^4\ \text{W}$$

A quarter of it is converted:

$$P = 0.25\times1.8\times10^4 = 4.5\times10^3\ \text{W}$$

Question 5.22

A person trying to lose weight (dieter) lifts a $10\ \text{kg}$ mass, one thousand times, to a height of $0.5\ \text{m}$ 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 $3.8\times10^7\ \text{J}$ of energy per kilogram which is converted to mechanical energy with a $20\%$ efficiency rate. How much fat will the dieter use up?

Question 5.22 (a)

Solution. Each lift raises the mass $0.5\ \text{m}$ against gravity and needs $mgh$ of work. Because the energy is dissipated each time she lowers it, none of it is recovered, and every one of the thousand lifts costs the full amount:

$$W = 1000\times10\times9.8\times0.5 = 4.9\times10^4\ \text{J}$$

Question 5.22 (b)

Solution. Only $20\%$ of the energy released from fat becomes mechanical work, so the body must release five times the work done:

$$E_{\text{fat}} = \frac{4.9\times10^4}{0.20} = 2.45\times10^5\ \text{J}, \qquad m_{\text{fat}} = \frac{2.45\times10^5}{3.8\times10^7} = 6.45\times10^{-3}\ \text{kg}$$

That is under $7\ \text{g}$ of fat for a thousand lifts, which is why exercise alone is a slow way to lose weight.

Question 5.23

A family uses $8\ \text{kW}$ of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of $200\ \text{W}$ per square meter. If $20\%$ of this energy can be converted to useful electrical energy, how large an area is needed to supply $8\ \text{kW}$? (b) Compare this area to that of the roof of a typical house.

Question 5.23 (a)

Solution. Each square metre receives $200\ \text{W}$, of which $20\%$, that is $40\ \text{W}$, becomes useful electrical power. The area needed is

$$A = \frac{8000\ \text{W}}{40\ \text{W m}^{-2}} = 200\ \text{m}^2$$

Question 5.23 (b)

Solution. $200\ \text{m}^2$ is a square about $14\ \text{m}$ on each side, since $\sqrt{200} \approx 14.1$. That is the roof of a large house. A typical house has a smaller roof than this, so it could meet only part of an $8\ \text{kW}$ demand from sunlight at this efficiency.

Common mistakes

• Question 5.1(d): calling the applied force’s work zero because the velocity is constant. Constant velocity means the net work is zero. The applied force still does positive work; friction does an equal amount of negative work, and the two cancel.
• Question 5.2: taking friction as $\mu\times7\ \text{N}$, or using $7\ \text{N}$ alone for the acceleration. Kinetic friction is $\mu$ times the normal reaction, which here is the weight, $mg$, not the applied force. The acceleration comes from the net force, $7 – 1.96 = 5.04\ \text{N}$.
• Question 5.5(d): counting “carrying” as work done on the mass. In everyday speech carrying a load is hard work, but the supporting force is vertical and the displacement horizontal, so $W = Fd\cos90^\circ = 0$. Pulling the rope in (ii) is along the motion and does $294\ \text{J}$.
• Question 5.7(a): believing each body keeps its own momentum in an elastic collision. Only the totals are conserved; the whole point of a collision is that the bodies exchange momentum and energy.
• Question 5.10: treating constant power as constant force. Constant force would give $x \propto t^2$. With constant power the force falls as the body speeds up, and energy gives $v \propto t^{1/2}$, so $x \propto t^{3/2}$.
• Question 5.13: assuming air resistance does no work in the first half, or forgetting the final kinetic energy. The resistance acts throughout the fall. The total work of all forces equals $\Delta K$, so $W_r = \tfrac12mv^2 – mgh$, about $-0.16\ \text{J}$, split almost equally between the two halves.
• Question 5.15: multiplying by the efficiency instead of dividing. The pump’s input is always larger than its useful output. $30\%$ efficiency means the input is the output divided by $0.30$.
• Question 5.19: expecting the trolley to speed up as it loses mass. That would be true only if the sand were thrown backwards, as a rocket’s exhaust is. Sand dropping through the floor leaves with the trolley’s own velocity and exerts no horizontal force.

Practise next

• Chapter 6, Systems of Particles and Rotational Motion — momentum conservation for many particles, and the rotational counterparts of work, kinetic energy and power.
• Chapter 4, Laws of Motion — friction and Newton’s laws, which the work calculations in 5.2 are built on.