Chapter 5: Work, Energy, and Power

Ans: Force exerted on the body, \[F=(-\hat{i}+2\hat{j}+3\hat{k})N\].

Displacement, \[s=4\hat{k}m\].

Hence the work done, \[W=F.s=(-\hat{i}+2\hat{j}+3\hat{k}).(4\hat{k})=12J\].

Therefore it is concluded that the work done is \[12J\].

Ans: The given mass of the electron, \[{{m}_{e}}=9.11\times {{10}^{-31}}kg\] and that of the proton is \[{{m}_{p}}=1.67\times {{10}^{-27}}kg\].

The given kinetic energy of the electron is \[{{T}_{e}}=10keV=1.60\times {{10}^{-15}}J\], and the kinetic energy of the proton is \[{{T}_{p}}=100keV=1.60\times {{10}^{-14}}J\].

The velocity of the electron is hence obtained as: \[{{v}_{e}}={{\left( \frac{2{{T}_{e}}}{{{m}_{e}}} \right)}^{1/2}}=5.93\times {{10}^{7}}m/s\]

Similarly, the velocity of the proton can be obtained as   \[{{v}_{p}}={{\left( \frac{2{{T}_{p}}}{{{m}_{p}}} \right)}^{1/2}}=4.38\times {{10}^{6}}m/s\]

The ratio of their velocities is \[\frac{{{v}_{e}}}{{{v}_{p}}}=\frac{13.54}{1}\]

Hence the electron moves 13.54 times faster.

Ans: The given radius of the raindrop, \[r=2\times {{10}^{-3}}m\]

The volume of the drop is\[V=(4/3)\pi {{r}^{3}}\]

And hence the mass is \[m=\rho V=\rho (4/3)\pi {{r}^{3}}\]

Therefore the gravitational force is \[F=mg=\rho (4/3)\pi {{r}^{3}}g\]

Thus the work by gravitation in the first half of the journey of the drop: \[{{W}_{1}}=F.s=0.082J\] and here \[s=250m\]

This is equal to the amount of work done in the second half of the journey, \[{{W}_{2}}=0.082J\]

If there is no resistance force, then the total energy of the raindrop will remain the same. Hence the total energy on the top is: \[{{E}_{T}}=mgh+0=0.164J\].

The resistive force makes the drop have a velocity of \[10m/s\]. Hence the total energy at the ground:

\[{{E}_{G}}=(1/2)m{{v}^{2}}=1.675\times {{10}^{-3}}J\]

Therefore the energy lost in the resistance is the difference between the energies: \[={{E}_{T}}-{{E}_{G}}=-0.162J\]

Hence the energy due to the resistive force is -0.162J.

Ans: The momentum of any system remains conserved as long as there aren’t any external forces acting on the system. Similarly, the momentum of the system in the gas container remains conserved irrespective of the collision being elastic or inelastic.

As the gas molecules strike the stationary wall of the container, they rebound at the same speed. Thus the velocity of the wall remains zero. This means that the collision is elastic. Hence, the total kinetic energy of the molecule remains conserved during the collision. 

Ans: Volume that can be filled in 15 minutes: \[V=30{{m}^{3}}\]

And the time of operation, \[t=15\min =900s\].

The tank is at \[h=40m\]height

And the given pump is of the efficiency \[\eta =30\%=0.3\].

The average power required for lifting the water to this height is:

\[P=\frac{E}{t}=\frac{mgh}{t}=13.067\times {{10}^{3}}W\]

This is the required power output, so we need a pump whose power output needs to be equal to this. In other words, the \[30\%\]of the pump's output should be equal to the required power.

\[P=30\%\,{{P}_{pump}}\]

\[\Rightarrow {{P}_{pump}}=\frac{100P}{30}=43.6\times {{10}^{3}}W=43.6kW\]

Hence we need a pump that consumes \[43.6kW\].

Ans: In all the cases of collisions where there is no external force, the linear momentum of the system is unchanged. In the case of elastic collision, the kinetic energy is also conserved.

Let’s say that all the ball bearings have mass- \[m\], so the total kinetic energy before collision:

\[{{T}_{i}}=\frac{1}{2}m{{V}^{2}}+\frac{1}{2}m{{0}^{2}}+\frac{1}{2}m{{0}^{2}}=\frac{1}{2}m{{V}^{2}}\]

Now, one needs to check that the final kinetic energy is equal to the initial one.

Case i)

The final kinetic energy \[{{T}_{f}}=\frac{1}{2}m{{0}^{2}}+\frac{1}{2}m{{\left( \frac{V}{2} \right)}^{2}}+\frac{1}{2}m{{\left( \frac{V}{2} \right)}^{2}}=\frac{1}{4}m{{V}^{2}}\].

Case ii)

The final kinetic energy is\[{{T}_{f}}=\frac{1}{2}m{{0}^{2}}+\frac{1}{2}m{{0}^{2}}+\frac{1}{2}m{{V}^{2}}=\frac{1}{2}m{{V}^{2}}\].

Case iii)

The final kinetic energy is \[{{T}_{f}}=\frac{1}{2}m{{\left( \frac{V}{3} \right)}^{2}}+\frac{1}{2}m{{\left( \frac{V}{3} \right)}^{2}}+\frac{1}{2}m{{\left( \frac{V}{3} \right)}^{2}}=\frac{1}{6}m{{V}^{2}}\]

Hence it is clearly seen that the kinetic energy is conserved only in case (ii).

Ans: If the collision is elastic, then the bob-A will transfer all its momentum to the other bob. After the collision, bob A will come to rest, and bob B will move with the initial velocity of A in the horizontal direction. Thus the bob A will not rise after the collision. 

Ans: As provided, the length of the pendulum, \[l=1.5m\] 

The mass of the bob \[=m\]

Energy dissipated \[=\text{ }5%\]

The law of conservation of momentum states the linear momentum of a system is unchanged in the absence of external force.

In the horizontal position

The PE of the bob is \[{{E}_{P}}=mgl\,\], while the KE is \[{{E}_{K}}=0\]and hence the total energy is \[E=mgl\]

While at the lowermost point or the mean position, the potential energy is \[{{E}_{P}}=0\], the KE is \[{{E}_{K}}=\frac{1}{2}m{{v}^{2}}\] , and the total energy is \[E=\frac{1}{2}m{{v}^{2}}\].

But the bob loses 5% of its energy. 

Hence the total energy at the lowermost point is 95% of the total energy at the higher/horizontal position

\[\Rightarrow \frac{1}{2}m{{v}^{2}}=\frac{95}{100}mgl\]

\[\Rightarrow v=5.28m/s\].

Hence the bob arrives with 5.28m/s at the lowest point.

Ans: As the trolley carrying the sand bag is moving without friction, which means, there is no friction which can act as an external force on the system. Hence the law of conservation of momentum is applicable.

The given initial mass of the bag \[{{m}_{s}}(t=0)=25kg\] and of the trolley is\[{{m}_{Tr}}=300kg\], hence the total initial mass is \[{{M}_{Tot}}(t=0)=325kg\].  And the mass drop rate is \[\frac{dM}{dt}=0.05kg/s\].

The initial velocity of the trolley is \[v(t=0)=\,27km/h=7.5m/s\]

The momentum at any given time \[t\]is \[p=(325-0.05t)v(t)\] and at \[t=0\], the momentum \[p(t=0)=325v(t=0)=325\times 27=8775kg\,\,m/s\]

The time at which the sandbag empties, \[{{t}_{final}}=\frac{25}{0.05}=500s\]

Thus the momentum at \[{{t}_{final}}\]can be written as  

\[8775=(325-0.05\times 500).v({{t}_{final}}=500)\]

\[\Rightarrow 8775=300.v({{t}_{final}})\]

\[\Rightarrow v({{t}_{final}})=29.25m/s\]

Hence the final velocity when the bag is emptied is \[29.25m/s\].

Ans: The given mass of the body is \[m=0.5kg\] and is moving with velocity\[v=a{{x}^{3/2}}\]where \[a=5{{m}^{1/2}}{{s}^{-1}}\].

The kinetic energy at \[x=0\] is \[{{E}_{K}}(x=0)=\frac{1}{2}0.5\times 0\] and at \[x=2\] is \[{{E}_{K}}(x=2)=\frac{1}{2}0.5\times {{(5\times {{2}^{3/2}})}^{2}}=50J\].

The difference in these two values is equal to the work done.

\[W={{E}_{K}}(x=2)-{{E}_{K}}(x=0)=50J\]

Hence the work done is \[50J\].