NCERT Exemplar Solutions Class 11 Physics Work, Energy and Power — Very Short and Short Answer Questions Q22–43

NCERT Exemplar Problems, Class 11 Physics — Work, Energy and Power, Very Short Answer Questions 22 to 33 and Short Answer Questions 34 to 43. All 22 questions solved. The book numbers them 6.22 to 6.43.

Three ideas carry this page. Work is force times the displacement of its point of application, so the same force can do work in one frame and none in another, and a force at right angles to the motion never does any. Energy bookkeeping: kinetic plus potential energy is conserved only when no friction or drag acts; when they do, the shortfall is the work done against them. Collisions conserve momentum always, kinetic energy only when elastic, and for two equal masses an elastic head-on collision simply swaps their velocities.

Key insight. Most of these questions ask where the energy went. Write the energy at the start, the energy at the end and name the force that accounts for the difference; Questions 26, 38, 41, 42 and 43 are each one line once that is done.

Very Short Answer Questions

Question 22

A rough inclined plane is placed on a cart moving with a constant velocity $u$ on horizontal ground. A block of mass $M$ rests on the incline. Is any work done by force of friction between the block and incline? Is there then a dissipation of energy?

Solution. In the ground frame the block moves horizontally with velocity $u$. The static friction on it acts up along the incline, which has a horizontal component along (or against) that motion, so friction does work on the block. The normal force does an equal and opposite amount of work, and gravity does none, so the net work is zero and the block’s speed stays constant. But there is no dissipation: the block does not slide over the incline, so there is no relative motion at the rough surface and no energy is turned into heat.

Question 23

Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case?

Solution. Left to itself, the loaded cage would fall freely under gravity. To make it descend at a steady, safe speed the motor must keep supplying an upward force on the cage through the cable, and producing that controlled force costs electrical power (and some is needed against friction in the mechanism). The more passengers there are, the larger the weight that must be held back, so the larger the force and power the motor must provide. The motor and brakes have a maximum force they can deliver, so the load must be limited, or the cage could not be stopped from accelerating downward.

Question 24

A body is being raised to a height $h$ from the surface of earth. What is the sign of work done by (a) applied force (b) gravitational force?

Solution. The applied force acts upward, along the upward displacement, so its work is positive. Gravity acts downward, against the displacement, so its work is negative, $-mgh$.

Question 25

Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is $400$ kg and the distance moved is $2$ m.

Solution. Gravity acts vertically and the displacement is horizontal, so the angle between them is $90^\circ$ and $W = Fd\cos 90^\circ = 0$. The mass and the distance do not matter.

Question 26

A body falls towards earth in air. Will its total mechanical energy be conserved during the fall? Justify.

Solution. No. Besides gravity, the air exerts a resistive drag on the body. Drag is a non-conservative force: it always opposes the motion and turns part of the mechanical energy into heat in the air and the body. So the gain in kinetic energy is less than the loss in potential energy.

Question 27

A body is moved along a closed loop. Is the work done in moving the body necessarily zero? If not, state the condition under which work done over a closed path is always zero.

Solution. No. Dragging a box round a closed loop on a rough floor takes work all the way round, because friction always opposes the motion and never gives the energy back. The work over every closed path is zero only when all the forces doing work are conservative, such as gravity or the force of an ideal spring. That is in fact the definition of a conservative force.

Question 28

In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls (i.e., when they are in contact)? (a) Kinetic energy. (b) Total linear momentum? Give reason for your answer in each case.

Solution.

• (a) Kinetic energy is not conserved during contact. While the balls are pressed together they deform, and part of the kinetic energy is stored as elastic potential energy. In an elastic collision it is all returned by the time they separate, so the K.E. after equals the K.E. before, but not during.
• (b) Total linear momentum is conserved at every instant. The only forces are the equal and opposite forces the balls exert on each other, which are internal to the pair.

Question 29

Calculate the power of a crane in watts, which lifts a mass of $100$ kg to a height of $10$ m in $20$ s.

Solution. The work done is the gain in potential energy, $mgh$, and power is work per unit time:

$$P = \frac{mgh}{t} = \frac{100 \times 9.8 \times 10}{20} = 490\ \text{W}.$$

With $g = 10\ \text{m s}^{-2}$ this would be $500$ W.

Question 30

The average work done by a human heart while it beats once is $0.5$ J. Calculate the power used by heart if it beats $72$ times in a minute.

Solution. In one minute the heart does $72 \times 0.5 = 36$ J of work, so

$$P = \frac{36\ \text{J}}{60\ \text{s}} = 0.6\ \text{W}.$$

Question 31

Give example of a situation in which an applied force does not result in a change in kinetic energy.

Solution. Any force that stays perpendicular to the velocity does no work, so it changes the direction of motion but not the speed. A charged particle moving in a uniform magnetic field is the classic case: the force $q\,\mathbf{v}\times\mathbf{B}$ bends it into a circle at constant speed. A stone whirled in a horizontal circle at constant speed on a string, where the tension is the centripetal force, is another.

Question 32

Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?

Solution. By the work-energy theorem, the work done by the retarding force equals the kinetic energy removed: $Fs = K$. Both bodies have the same $K$ and feel the same $F$, so they stop in the same distance $s = K/F$. Their masses do not enter. (The lighter one is moving faster, but it also decelerates faster, and the two effects cancel exactly.)

Question 33

A bob of mass $m$ suspended by a light string of length $L$ is whirled into a vertical circle as shown in Fig. 6.11. What will be the trajectory of the particle if the string is cut at (a) Point B? (b) Point C? (c) Point X?

In Fig. 6.11, A is the lowest point of the circle, C the highest, B the point level with the centre on the right, and X a point on the upper left, between the left end of the horizontal diameter and C. The arrows show the bob moving up the left side and down the right: from A through X to C and then down through B.

Solution. When the string is cut, the bob becomes a projectile that starts with its velocity at that instant, which is along the tangent to the circle.

• (a) At B the bob is moving vertically downward. Its velocity and gravity are both vertical, so it falls along a vertical straight line, downward.
• (b) At C the bob is moving horizontally. It is projected horizontally, so it follows a parabola whose vertex is at C.
• (c) At X the velocity is along the tangent, pointing up and to the right. The bob follows a parabola whose vertex is higher than C. To see why, compare energies: at the vertex only the horizontal part of the velocity remains, $v_X\cos\phi$, where $\phi$ is the angle of X from the top. Using $v_X^2 = v_C^2 + 2gL(1-\cos\phi)$ and $v_C^2 \ge gL$ (needed to complete the circle), one finds $v_X\cos\phi < v_C$, so the bob keeps more potential energy at its vertex than it has at C.

Short Answer Questions

Question 34

A graph of potential energy $V(x)$ verses $x$ is shown in Fig. 6.12. A particle of energy $E_0$ is executing motion in it. Draw graph of velocity and kinetic energy versus $x$ for one complete cycle AFA.

From the figure: the particle is confined between a vertical wall at $x = 0$ (where the potential jumps from B up past A) and the sloping wall DF, which reaches $E_0$ at F. Between them the potential has a small hump just after B, falls to zero at C, stays zero from C to D, and rises linearly from D to F.

Solution. Everything follows from $K(x) = E_0 – V(x)$ and $v = \pm\sqrt{2K/m}$.

• Kinetic energy. At $x = 0$ the particle has $K = E_0 – V_B$. Over the small hump $K$ dips slightly, then it rises to $E_0$ at C, stays at $E_0$ from C to D where $V = 0$, and falls linearly to zero at F, the turning point. Since $K$ depends only on $x$, the curve is the same on the way out (A to F) and on the way back (F to A).
• Velocity. On the way out $v$ is positive, $v = +\sqrt{2K/m}$; on the way back it is negative. Between C and D the speed is constant at $\sqrt{2E_0/m}$. Near F, $K$ falls linearly, so $v \propto \sqrt{x_F – x}$, and the curve meets the $x$-axis at F with a vertical tangent: the particle stops and turns round smoothly. At $x = 0$ the vertical wall reverses the velocity abruptly, from $-\sqrt{2K_B/m}$ to $+\sqrt{2K_B/m}$, which closes the loop. So the $v$-$x$ graph is a closed curve, symmetric about the $x$-axis, traced clockwise.

Question 35

A ball of mass $m$, moving with a speed $2v_0$, collides inelastically $(e > 0)$ with an identical ball at rest. Show that (a) For head-on collision, both the balls move forward. (b) For a general collision, the angle between the two velocities of scattered balls is less than $90^\circ$.

Solution.

(a) Let the velocities after the collision be $v_1$ and $v_2$ along the original direction. Momentum conservation and the definition of the coefficient of restitution give

$$m(2v_0) = mv_1 + mv_2 \;\Rightarrow\; v_1 + v_2 = 2v_0, \qquad e = \frac{v_2 – v_1}{2v_0} \;\Rightarrow\; v_2 – v_1 = 2ev_0.$$

Solving, $v_1 = v_0(1 – e)$ and $v_2 = v_0(1 + e)$. The collision is inelastic, so $e < 1$, and $v_1 > 0$; $v_2$ is plainly positive. Both balls move forward.

(b) Let the momenta be $\mathbf{p}$ before and $\mathbf{p}_1$, $\mathbf{p}_2$ after, so $\mathbf{p} = \mathbf{p}_1 + \mathbf{p}_2$. Kinetic energy is lost in an inelastic collision, and the masses are equal, so

$$\frac{p^2}{2m} > \frac{p_1^2}{2m} + \frac{p_2^2}{2m} \;\Rightarrow\; p^2 > p_1^2 + p_2^2.$$

But $p^2 = |\mathbf{p}_1 + \mathbf{p}_2|^2 = p_1^2 + p_2^2 + 2p_1p_2\cos\theta$, where $\theta$ is the angle between the two outgoing velocities. Hence $2p_1p_2\cos\theta > 0$, so $\cos\theta > 0$ and $\theta < 90^\circ$. (For an elastic collision the inequality becomes an equality and $\theta = 90^\circ$ exactly, the familiar billiard-ball result.)

Question 36

Consider a one-dimensional motion of a particle with total energy $E$. There are four regions A, B, C and D in which the relation between potential energy $V$, kinetic energy $(K)$ and total energy $E$ is as given below:

Region A: $V > E$. Region B: $V < E$. Region C: $K > E$. Region D: $V > K$.

State with reason in each case whether a particle can be found in the given region or not.

Solution. Use $K = E – V$ and the fact that $K = \tfrac12 mv^2$ can never be negative.

• Region A, $V > E$: not possible. It would make $K = E – V$ negative.
• Region B, $V < E$: possible. Then $K = E – V$ is positive; the particle is simply moving.
• Region C, $K > E$: possible, provided the potential energy there is negative, since $K = E – V > E$ means $V < 0$. Potential energy has an arbitrary zero and can be negative, as near a planet.
• Region D, $V > K$: possible. Nothing forbids the potential energy exceeding the kinetic energy, for example near a turning point where $K$ is small.

Question 37

The bob A of a pendulum released from horizontal to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.13. If the length of the pendulum is $1$ m, calculate (a) the height to which bob A will rise after collision. (b) the speed with which bob B starts moving. Neglect the size of the bobs and assume the collision to be elastic.

The figure shows bob A held with the string horizontal, $1$ m above the table, and bob B resting on the table directly below the point of suspension, where A reaches the bottom of its swing.

Solution. A falls through $h = 1$ m, so just before the collision its speed is $\sqrt{2gh}$. The two bobs have equal mass and the collision is elastic and head-on, so they exchange velocities.

• (a) A comes to rest and does not rise at all: height $0$.
• (b) B moves off with A’s speed: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1} = 4.43\ \text{m/s}$.

Question 38

A raindrop of mass $1.00$ g falling from a height of $1$ km hits the ground with a speed of $50\ \text{m s}^{-1}$. Calculate (a) the loss of P.E. of the drop. (b) the gain in K.E. of the drop. (c) Is the gain in K.E. equal to loss of P.E.? If not why. Take $g = 10\ \text{m s}^{-2}$.

Solution.

$$\text{(a)}\ \ \Delta U = mgh = 1.00\times10^{-3} \times 10 \times 10^3 = 10\ \text{J}.$$

$$\text{(b)}\ \ \Delta K = \tfrac12 mv^2 = \tfrac12 \times 1.00\times10^{-3} \times 50^2 = 1.25\ \text{J}.$$

(c) No. Only $1.25$ J of the $10$ J appears as kinetic energy. The remaining $8.75$ J is used up in doing work against the viscous drag of the air, which is a non-conservative force; it goes into heating the air and the drop.

Question 39

Two pendulums with identical bobs and lengths are suspended from a common support such that in rest position the two bobs are in contact (Fig. 6.14). One of the bobs is released after being displaced by $10^\circ$ so that it collides elastically head-on with the other bob. (a) Describe the motion of two bobs. (b) Draw a graph showing variation in energy of either pendulum with time, for $0 \le t \le 2T$, where $T$ is the period of each pendulum.

Solution.

(a) Say bob A is released from $10^\circ$ at $t = 0$. It reaches the bottom after a quarter period, $t = T/4$, and hits B. Equal masses in an elastic head-on collision exchange velocities, so A stops dead and B swings out to $10^\circ$ on the other side. B returns to the bottom after half a period, at $t = 3T/4$, hits A, stops, and A swings out and back, returning at $t = 5T/4$. The bobs take turns: at any time one is swinging and the other is at rest, and the energy passes back and forth at every collision.

(b) Between collisions each pendulum’s own energy (kinetic plus potential) is constant. So the energy $E_A$ of A is $E_0$ from $0$ to $T/4$, zero from $T/4$ to $3T/4$, $E_0$ from $3T/4$ to $5T/4$, zero from $5T/4$ to $7T/4$, and $E_0$ again from $7T/4$ to $2T$. The energy $E_B$ of B is the mirror image: zero whenever $E_A = E_0$, and $E_0$ whenever $E_A$ is zero. Their sum is $E_0$ throughout.

Question 40

Suppose the average mass of raindrops is $3.0 \times 10^{-5}$ kg and their average terminal velocity $9\ \text{m s}^{-1}$. Calculate the energy transferred by rain to each square metre of the surface at a place which receives $100$ cm of rain in a year.

Solution. $100$ cm of rain on $1\ \text{m}^2$ is a volume of $1\ \text{m}^2 \times 1\ \text{m} = 1\ \text{m}^3$ of water, of mass $M = 10^3\ \text{kg}$. All of it arrives at the terminal speed of $9$ m/s, so the energy delivered is

$$E = \tfrac12 Mv^2 = \tfrac12 \times 10^3 \times 9^2 = 4.05 \times 10^4\ \text{J}.$$

The mass of a single drop is not needed: it only tells us that about $3.3 \times 10^7$ drops fall on each square metre.

Question 41

An engine is attached to a wagon through a shock absorber of length $1.5$ m. The system with a total mass of $50{,}000$ kg is moving with a speed of $36\ \text{km h}^{-1}$ when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by $1.0$ m. If $90\%$ of energy of the wagon is lost due to friction, calculate the spring constant.

Solution. $36$ km/h $= 10$ m/s, so the kinetic energy of the system is

$$K = \tfrac12 mv^2 = \tfrac12 \times 5\times10^4 \times 10^2 = 2.5 \times 10^6\ \text{J}.$$

$90\%$ is lost to friction, so $10\%$, that is $2.5 \times 10^5$ J, is stored in the spring:

$$\tfrac12 kx^2 = 2.5\times10^5 \;\Rightarrow\; k = \frac{2 \times 2.5\times10^5}{(1.0)^2} = 5 \times 10^5\ \text{N/m}.$$

Printed key error. The key writes $\tfrac12 \times 5\times10^4 \times 10^2 = 2.5 \times 10^5$ J, dropping a factor of $10$ (the product is $2.5 \times 10^6$ J). Carrying that slip through, it stores $2.5 \times 10^4$ J in the spring and gets $k = 5 \times 10^4$ N/m. The correct spring constant is $5 \times 10^5$ N/m.

Question 42

An adult weighing $600$ N raises the centre of gravity of his body by $0.25$ m while taking each step of $1$ m length in jogging. If he jogs for $6$ km, calculate the energy utilised by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting $10\%$ of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilised for jogging.

Solution. In $6$ km there are $6000$ steps of $1$ m. Each step lifts his weight through $0.25$ m, costing $600 \times 0.25 = 150$ J, so

$$E = 6000 \times 150 = 9 \times 10^5\ \text{J}.$$

The body turns only $10\%$ of food energy into this work, so the food needed is ten times as much: $9 \times 10^6$ J.

Question 43

On complete combustion a litre of petrol gives off heat equivalent to $3 \times 10^7$ J. In a test drive a car weighing $1200$ kg, including the mass of driver, runs $15$ km per litre while moving with a uniform speed on a straight track. Assuming that friction offered by the road surface and air to be uniform, calculate the force of friction acting on the car during the test drive, if the efficiency of the car engine were $0.5$.

Solution. With efficiency $0.5$, one litre provides $0.5 \times 3\times10^7 = 1.5 \times 10^7$ J of useful work. The car moves at uniform speed, so its kinetic energy does not change and all of this work is done against friction over $15$ km:

$$F \times 15\,000 = 1.5 \times 10^7 \;\Rightarrow\; F = 1000\ \text{N}.$$

The mass of the car is not needed.

Common mistakes

• Saying friction does no work in Question 22 because the block does not slide. It does work in the ground frame; what it does not do is dissipate energy.
• Mixing up “during” and “after” in Question 28. An elastic collision restores the kinetic energy at the end, but during contact some of it is elastic potential energy.
• Making the bob fly off horizontally at B in Question 33. At B the velocity is vertical, so the path is a vertical line.
• Letting K.E. depend on direction in Question 34. $K = E_0 – V(x)$ is the same on the way out and back; only the sign of $v$ changes.
• Making both bobs move after the collision in Questions 37 and 39. Equal masses in an elastic head-on collision exchange velocities: the striker stops.
• Using the mass of one raindrop in Question 40. The energy depends on the total mass of rain, $10^3$ kg per square metre.
• Dropping a power of ten in Question 41. $\tfrac12 \times 5\times10^4 \times 100$ is $2.5 \times 10^6$ J, which gives $k = 5 \times 10^5$ N/m.

Practise next

• A pump lifts $500$ kg of water to a tank $20$ m above it in $50$ s. Find the power it delivers. Take $g = 9.8\ \text{m s}^{-2}$.

Show answer

$1960$ W. The work is $mgh = 500 \times 9.8 \times 20 = 98\,000$ J, and $98\,000/50 = 1960$ W.

• A ball moving at $6$ m/s collides head-on with an identical ball at rest. The coefficient of restitution is $0.5$. Find both velocities after the collision.

Show answer

$1.5$ m/s and $4.5$ m/s, both forward. With $2v_0 = 6$ m/s, $v_1 = v_0(1-e) = 3 \times 0.5 = 1.5$ m/s and $v_2 = v_0(1+e) = 3 \times 1.5 = 4.5$ m/s; momentum $1.5 + 4.5 = 6$ checks.

• Exemplar Work, Energy and Power: Multiple Choice Questions — the same ideas in objective form.
• Exemplar Work, Energy and Power: Long Answer Questions — the incline with friction, rolling balls, the rocket, colliding cubes and the rising balloon.
• NCERT textbook Chapter 5 — Work, Energy and Power for the theory of work, energy, power and collisions.