NCERT Exemplar Solutions Class 11 Physics Work, Energy and Power — Long Answer Questions Q44–48

NCERT Exemplar Problems, Class 11 Physics — Work, Energy and Power, Long Answer Questions 44 to 48. All 5 questions solved. The book numbers them 6.44 to 6.48; here they are Questions 44 to 48.

Three ideas carry this page. The work-energy theorem with every force counted: the work of the applied force is shared between raising potential energy, overcoming friction and adding kinetic energy. Energy can hide in places that are easy to miss: in the rotation of a rolling ball, in the exhaust gas of a rocket, in the elastic compression of a steel block, and in the air a balloon pushes down as it rises. Mechanical energy is conserved for the whole system, not for one part of it, when only conservative forces act.

Key insight. When energy seems to appear from nowhere or vanish, widen the system. The rising balloon of Question 48 gains energy because the air it displaces falls, and the rocket of Question 46 gains energy because its gas-ejecting device does work.

Long Answer Questions

Question 44

A block of mass $1$ kg is pushed up a surface inclined to horizontal at an angle of $30^\circ$ by a force of $10$ N parallel to the inclined surface (Fig. 6.15). The coefficient of friction between block and the incline is $0.1$. If the block is pushed up by $10$ m along the incline, calculate (a) work done against gravity (b) work done against force of friction (c) increase in potential energy (d) increase in kinetic energy (e) work done by applied force.

The figure shows the block on a $30^\circ$ incline with the force $F$ acting up the slope, parallel to it. Take $g = 10\ \text{m s}^{-2}$.

Solution. Along the incline the block moves $d = 10$ m, rising $h = d\sin30^\circ = 5$ m.

• (a) Work against gravity: $W_g = mg\sin\theta \cdot d = 1 \times 10 \times 0.5 \times 10 = 50$ J.
• (b) Friction is $\mu N = \mu mg\cos\theta$, so $W_f = 0.1 \times 1 \times 10 \times 0.866 \times 10 = 8.66$ J.
• (c) The increase in potential energy is the work done against gravity: $\Delta U = mgh = 1 \times 10 \times 5 = 50$ J.
• (e) Work by the applied force: $W_F = Fd = 10 \times 10 = 100$ J.
• (d) By the work-energy theorem, the kinetic energy gained is what is left over:

$$\Delta K = W_F – W_g – W_f = 100 – 50 – 8.66 = 41.34\ \text{J}.$$

As a check, the net force is $10 – 5 – 0.866 = 4.13$ N, so $a = 4.13\ \text{m s}^{-2}$ and $\Delta K = mad = 1 \times 4.13 \times 10 = 41.3$ J.

Question 45

A curved surface is shown in Fig. 6.16. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from rest one by one from A which is at a slightly greater height than C.

The figure shows a track that falls from A to a valley at B, rises to a hump at C and falls again towards D. A is only slightly higher than C.

With the surface AB, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has a small friction and ball 3 has a negligible friction. (a) For which balls is total mechanical energy conserved? (b) Which ball(s) can reach D? (c) For balls which do not reach D, which of the balls can reach back A?

Solution.

(a) Ball 1 rolls without slipping, so its point of contact is momentarily at rest and static friction does no work: mechanical energy (translational plus rotational) is conserved. Ball 3 slides on a frictionless surface, so its energy is conserved too. Ball 2 slips while friction acts, and sliding friction turns some energy into heat. Balls 1 and 3.

(b) To get over C a ball needs enough translational kinetic energy, because on the frictionless part BCD nothing can change its spin, and the spin energy cannot be used to climb. Ball 3 does not spin, so all of $mgh_A$ is translational and, since A is higher than C, it crosses C and reaches D. Ball 1 has put $\tfrac{2}{7}$ of its energy into rotation (for a solid sphere $K_{\text{rot}}/K = \tfrac{2}{7}$), leaving only $\tfrac{5}{7}mgh_A$ for climbing, which is less than $mgh_C$ when A is only slightly above C. Ball 2 has lost energy to friction and also spins. Only ball 3 reaches D.

(c) Balls 1 and 2 turn back before C. Ball 2 cannot reach A because it has lost energy. Ball 1 is subtler: on BCD its spin is unchanged, so when it comes back to B it is still spinning in the sense it had while rolling forward, the “wrong” sense for rolling back up AB. On AB it therefore slips, and kinetic friction dissipates energy, so it cannot reach A either. Neither ball can reach back A.

Question 46

A rocket accelerates straight up by ejecting gas downwards. In a small time interval $\Delta t$, it ejects a gas of mass $\Delta m$ at a relative speed $u$. Calculate KE of the entire system at $t + \Delta t$ and $t$ and show that the device that ejects gas does work $= \left(\tfrac12\right)\Delta m\,u^2$ in this time interval (neglect gravity).

Solution. At time $t$ the rocket (with the gas still inside) has mass $M$ and velocity $v$, so

$$K_t = \tfrac12 Mv^2.$$

At $t + \Delta t$ the rocket of mass $M – \Delta m$ moves at $v + \Delta v$, and the ejected gas moves at $v – u$:

$$K_{t+\Delta t} = \tfrac12 (M – \Delta m)(v + \Delta v)^2 + \tfrac12 \Delta m (v – u)^2.$$

With gravity neglected there is no external force, so momentum is conserved:

$$Mv = (M – \Delta m)(v + \Delta v) + \Delta m (v – u) \;\Rightarrow\; (M – \Delta m)\Delta v = \Delta m\,u.$$

Expanding the kinetic energy and keeping first-order terms,

$$K_{t+\Delta t} – K_t = (M – \Delta m)v\Delta v – \Delta m\,vu + \tfrac12 \Delta m\,u^2 + \tfrac12 (M – \Delta m)(\Delta v)^2.$$

The first two terms cancel by the momentum equation, $(M – \Delta m)\Delta v = \Delta m\,u$. The last term is of second order, since $\Delta v$ is itself proportional to $\Delta m$. So

$$\Delta K = \tfrac12 \Delta m\,u^2.$$

(Kept exactly, $\Delta K = \dfrac{M\,\Delta m\,u^2}{2(M – \Delta m)}$, which tends to $\tfrac12\Delta m\,u^2$ as $\Delta m \to 0$.) No external force acts, so by the work-energy theorem this gain in kinetic energy must be supplied by the internal device that ejects the gas: it does work $W = \tfrac12 \Delta m\,u^2$.

Question 47

Two identical steel cubes (masses $50$ g, side $1$ cm) collide head-on face to face with a speed of $10$ cm/s each. Find the maximum compression of each. Young’s modulus for steel $= Y = 2 \times 10^{11}\ \text{N/m}^2$.

Solution. Treat each cube as a spring. By Hooke’s law $F/A = Y\,\Delta L/L$, so $F = (YA/L)\Delta L$, and the spring constant is

$$k = \frac{YA}{L} = \frac{YL^2}{L} = YL = 2\times10^{11} \times 0.01 = 2 \times 10^9\ \text{N/m}.$$

The cubes approach with equal and opposite momenta, so at maximum compression both are momentarily at rest and all the kinetic energy is stored elastically:

$$K = 2 \times \tfrac12 mv^2 = 2 \times \tfrac12 \times 0.05 \times (0.1)^2 = 5 \times 10^{-4}\ \text{J}.$$

Each cube is compressed by the same $\Delta L$ and stores $\tfrac12 k(\Delta L)^2$, so

$$2 \times \tfrac12 k(\Delta L)^2 = K \;\Rightarrow\; \Delta L = \sqrt{\frac{K}{k}} = \sqrt{\frac{5\times10^{-4}}{2\times10^{9}}} = \sqrt{2.5\times10^{-13}} = 5.0 \times 10^{-7}\ \text{m}.$$

Printed key error. The key sets up exactly this calculation, with $k = YL$, but substitutes $L = 0.1$ m in the last line, $\sqrt{5\times10^{-4}/(2\times10^{11}\times0.1)}$, and gets $1.58 \times 10^{-7}$ m. The cube’s side is $1$ cm $= 0.01$ m, which gives $5.0 \times 10^{-7}$ m.

Question 48

A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that density of air is constant.

Solution. The balloon alone is not an isolated system; the air around it must be counted. Let the balloon have mass $m$ and volume $V$, with helium density $\rho_{He}$ and air density $\rho_{air}$. The net upward force is the buoyancy minus the weight of the helium (taking the balloon’s mass to be that of its helium):

$$V(\rho_{air} – \rho_{He})g = ma.$$

Starting from rest, after time $t$ it has $v = at$ and has risen $h = \tfrac12 at^2$, so

$$\tfrac12 mv^2 = \tfrac12 m a^2t^2 = ma \cdot h = V(\rho_{air} – \rho_{He})gh.$$

Rearranging,

$$\underbrace{\tfrac12 mv^2}_{\text{K.E. of balloon}} + \underbrace{V\rho_{He}\,gh}_{\text{P.E. gained by balloon}} = \underbrace{V\rho_{air}\,gh}_{\text{P.E. lost by air}}.$$

As the balloon rises through $h$, an equal volume $V$ of air moves down through $h$ to fill the space it leaves. That air loses potential energy $V\rho_{air}gh$, and this is exactly what pays for both the kinetic energy and the potential energy that the balloon gains. The mechanical energy of the whole system, balloon plus air, is conserved.

Common mistakes

• Forgetting friction when finding the kinetic energy in Question 44. The applied force’s $100$ J is shared three ways: $50$ J to gravity, $8.66$ J to friction and $41.3$ J to kinetic energy.
• Thinking the rolling ball can climb as high as it started in Question 45. On the frictionless hump its spin energy is locked in and cannot help it rise.
• Letting ball 1 roll back to A in Question 45. It returns spinning the wrong way, slips and loses energy.
• Treating the exhaust gas as having no kinetic energy in Question 46. The gas moves at $v – u$, and its K.E. must be counted.
• Putting in $L = 0.1$ m for a $1$ cm cube in Question 47. That is the slip in the printed key; $k = YL$ with $L = 0.01$ m gives $5.0 \times 10^{-7}$ m.
• Applying energy conservation to the balloon alone in Question 48. The buoyant force comes from the air, and the air’s falling is what supplies the energy.

Practise next

• A $2$ kg block is pushed $5$ m up a $30^\circ$ incline by a force of $20$ N parallel to the incline. The coefficient of friction is $0.2$ and $g = 10\ \text{m s}^{-2}$. Find the work done against gravity, the work done against friction and the gain in kinetic energy.

• Two identical steel cubes of side $2$ cm and mass $64$ g collide head-on, face to face, each moving at $0.25$ m/s. Taking $Y = 2 \times 10^{11}\ \text{N/m}^2$, find the maximum compression of each cube.

• Exemplar Work, Energy and Power: Very Short and Short Answer Questions — power, collisions and energy graphs.
• Exemplar Work, Energy and Power: Multiple Choice Questions — the same ideas in objective form.
• NCERT textbook Chapter 5 — Work, Energy and Power for the theory of the work-energy theorem, potential energy and collisions.
• Exemplar Systems of Particles and Rotational Motion — the next unit, where the rotational energy of Question 45 is treated in full.