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

Work is positive when the force has a component along displacement, and negative when it opposes displacement. The man's pull lifts the bucket upward, so it does positive work. Gravity acts downward while the bucket moves upward, so its work is negative. Friction and air resistance oppose motion, so their work is negative. For uniform motion on a rough horizontal plane, the applied force balances friction and acts along displacement, so its work is positive.

Friction is $f_k=\mu mg=0.1\times2\times10=2\,\text{N}$. Net force $=7-2=5\,\text{N}$, so $a=5/2=2.5\,\text{m s}^{-2}$. From rest in 10 s, $s=\frac12at^2=\frac12(2.5)(100)=125\,\text{m}$. Applied work $=7\times125=875\,\text{J}$; friction work $=-2\times125=-250\,\text{J}$; net work $=625\,\text{J}$. Final speed $v=at=25\,\text{m s}^{-1}$, so $\Delta K=\frac12(2)(25)^2=625\,\text{J}$, equal to the net work.

$V(x)=\frac12kx^2=\frac12(0.5)x^2=0.25x^2$. At $x=\pm2\,\text{m}$, $V=0.25(4)=1\,\text{J}$, equal to the total energy. Hence kinetic energy $K=E-V=0$ at these points. Beyond them $V>E$, which is impossible, so the particle turns back.

Frictional heating of the casing comes from the rocket's mechanical energy. Gravitational force is conservative, so its work over a complete closed path is zero. For an approximately circular orbit, $v=\sqrt{GM/r}$; as drag lowers the satellite to smaller $r$, its speed increases even though its total energy becomes more negative. In Fig. 5.13(i), the force by the man on the load is vertical while displacement is horizontal, so work on the load is zero. In Fig. 5.13(ii), the hanging mass rises by 2 m, so work is $mgh=15\times10\times2=300\,\text{J}$.

For a conservative force, $W=-\Delta V$, so positive work means potential energy decreases. Work against friction is dissipative and normally reduces mechanical kinetic energy. Internal forces cancel in the total momentum equation, so $d\vec P/dt$ is set by the external force. In an isolated inelastic collision, total linear momentum is conserved and total energy is conserved including internal/thermal forms, but kinetic energy is not conserved.

During contact in an elastic collision, some kinetic energy is temporarily stored as elastic potential energy of deformation, so kinetic energy need not be constant at every instant, though it is the same before and after the collision. Momentum is conserved throughout because the impulsive forces are internal and external impulse is negligible. In an inelastic collision, kinetic energy is not restored after collision, but momentum conservation still holds for the isolated pair. If the interaction potential depends only on separation, the force is conservative, so the collision is elastic.

1. i. t^(1/2)
2. ii. t
3. iii. t^(3/2)
4. iv. t^2

With constant acceleration from rest, $v=at$. The force is constant, $F=ma$. Power is $P=Fv=ma(at)=ma^2t$, so $P\propto t$.

1. i. t^(1/2)
2. ii. t
3. iii. t^(3/2)
4. iv. t^2

Constant power gives $P=Fv=m(dv/dt)v$. Thus $v\,dv=(P/m)dt$, so $v^2\propto t$ and $v\propto t^{1/2}$. Since $v=dx/dt$, displacement $x\propto\int t^{1/2}dt\propto t^{3/2}$.

The displacement is $\vec s=4\hat{k}\,\text{m}$. Work is the dot product $W=\vec F\cdot\vec s=(-\hat{i}+2\hat{j}+3\hat{k})\cdot(4\hat{k})=12\,\text{J}$.

For non-relativistic particles, $K=\frac12mv^2$, so $v\propto\sqrt{K/m}$. Therefore $\frac{v_e}{v_p}=\sqrt{\frac{K_e m_p}{K_p m_e}}=\sqrt{\frac{10}{100}\times\frac{1.67\times10^{-27}}{9.11\times10^{-31}}}=\sqrt{183.3}=13.5$. Hence the electron is faster.

Mass of the drop is $m=\rho\frac{4}{3}\pi r^3=1000\times\frac{4}{3}\pi(2\times10^{-3})^3=3.35\times10^{-5}\,\text{kg}$. Work by gravity over each 250 m half is $mgh=3.35\times10^{-5}\times10\times250=8.38\times10^{-2}\,\text{J}$. Total work by gravity is $1.68\times10^{-1}\,\text{J}$. The final kinetic energy is $\frac12mv^2=\frac12(3.35\times10^{-5})(10)^2=1.68\times10^{-3}\,\text{J}$. By work-energy theorem, $W_g+W_r=\Delta K$, so $W_r=1.68\times10^{-3}-1.68\times10^{-1}\approx-1.66\times10^{-1}\,\text{J}$.

The molecule reverses the component of velocity normal to the wall, so its momentum changes; the wall receives an equal and opposite impulse. Therefore total momentum of the molecule plus wall is conserved. Since the molecule rebounds with the same speed, its kinetic energy is unchanged, so the collision is elastic.

Mass of water is $m=\rho V=1000\times30=3.0\times10^4\,\text{kg}$. Useful work is $mgh=3.0\times10^4\times10\times40=1.2\times10^7\,\text{J}$. Time $=15\,\text{min}=900\,\text{s}$, so useful power is $1.2\times10^7/900=1.33\times10^4\,\text{W}$. With efficiency 0.30, electric input power is $1.33\times10^4/0.30=4.44\times10^4\,\text{W}$.

1. i. Ball 1 stops and balls 2 and 3 move together with speed V/2.
2. ii. Balls 1 and 2 stop and ball 3 moves with speed V.
3. iii. All three balls move together with speed V/3.

Initial momentum is $mV$ and initial kinetic energy is $\frac12mV^2$. In option (ii), final momentum is $mV$ and final kinetic energy is $\frac12mV^2$, so both are conserved. Options (i) and (iii) conserve momentum but give smaller final kinetic energy, so they cannot be the result of an elastic collision.

At the lowest point, bob A collides elastically and head-on with bob B of equal mass initially at rest. In such a collision, the moving body transfers its velocity to the resting body and comes to rest. Therefore bob A has no kinetic energy immediately after collision and does not rise.

From the horizontal position to the lowest point, the bob falls through height $L=1.5\,\text{m}$. Only 95% of the initial potential energy becomes kinetic energy: $\frac12mv^2=0.95mgL$. Thus $v=\sqrt{2(0.95)(10)(1.5)}=\sqrt{28.5}=5.34\,\text{m s}^{-1}$.

The sand leaks out vertically relative to the trolley and has the same horizontal speed as the trolley at the instant it leaves. With no external horizontal force and no relative horizontal ejection speed, the trolley experiences no horizontal thrust. Therefore its speed remains $27\,\text{km h}^{-1}$.

By the work-energy theorem, $W=\Delta K$. Since $v=ax^{3/2}$, at $x=0$, $v=0$; at $x=2\,\text{m}$, $v=5(2)^{3/2}\,\text{m s}^{-1}$, so $v^2=25(2^3)=200\,\text{m}^2\text{s}^{-2}$. Thus $W=\frac12mv^2=\frac12(0.5)(200)=50\,\text{J}$.

In time $t$, air in a cylinder of volume $Avt$ crosses the blades, so mass $m=\rho Avt$. Its kinetic energy is $K=\frac12mv^2=\frac12\rho Av^3t$. Electrical power at 25% efficiency is $P=0.25\times\frac12\rho Av^3$. With $v=36\,\text{km h}^{-1}=10\,\text{m s}^{-1}$, $P=0.125(1.2)(30)(10^3)=4500\,\text{W}=4.5\,\text{kW}$.

Work per lift is $mgh=10\times10\times0.5=50\,\text{J}$. For 1000 lifts, $W=5.0\times10^4\,\text{J}$. Usable mechanical energy from 1 kg fat is $0.20(3.8\times10^7)=7.6\times10^6\,\text{J}$. Fat used $=5.0\times10^4/(7.6\times10^6)=6.6\times10^{-3}\,\text{kg}$.

Useful solar power per square metre is $0.20\times200=40\,\text{W m}^{-2}$. Required area $A=8000/40=200\,\text{m}^2$. A typical house roof is often of the order of $100$ to $200\,\text{m}^2$, so the required area is roughly a whole roof or somewhat more.