Work done on a body is reflected as change in kinetic energy, according to Work-Energy theorem.
\(\mathrm{W}=\int \mathrm{F} d x=\int m \frac{d v}{d f} d x\)
\(=\int m v d v=\left|\frac{1}{2} m v^{2}\right|_{v i}^{v_{f}}\)
Work done \(=\frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)\)
\(=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}\)
Work done = Change in kinetic energy.

At A
P.E. mgh K.E. = 0
Total energy = mgh

At B
\(\text { Velocity, },=\sqrt{2 g h}\)
\(\mathrm{K.E.}=\frac{1}{2} m v^{2}=\frac{1}{2} m \cdot 2 g x=m g x\)
P.E. = mg (h -x)
Total energy = mgh.
At C  
\(\text { Velocity, },=\sqrt{2 g h}\)
\(\mathrm{K.E.}=\frac{1}{2} m v^{2}=\frac{1}{2} m \cdot 2 g h=m g h\)
P.E. = 0
ஃ total energy = mgh.
Total mechanical energy is therefore mgh at all states as a body is dropped.

Graphs depicting variation of (i) gravitational potential energy (P.E.), (ii) kinetic energy (K.E.),and (iii) the total sum of potential and kinetic energies for a freely falling body are as shown in adjoining Fig. From the graphs, it is clear that:
(a) Gravitational potential energy decreases as the body falls downwards and is zero at the Earth.

(b) Kinetic energy increases as the body falls downwards and is maximum when the body just strikes the ground.
(c) The sum of kinetic and potential energies remains constant at all points during its free fall.

\({ K }_{ 1 }=\frac { F }{ { x }_{ 1 } } and\ { K }_{ 2 }=\frac { F }{ { x }_{ 2 } } \)
Since K₁ > K₂
\(\therefore\) x₁ < x₂
\({ W }_{ 1 }=\frac { 1 }{ 2 } { K }_{ 1 }{ x }_{ 1 }^{ 2 }\)  and \({ W }_{ 2 }=\frac { 1 }{ 2 } { K }_{ 2 }{ x }_{ 2 }^{ 2 }\)
\(\frac { { W }_{ 1 } }{ { W }_{ 2 } } =\frac { \frac { 1 }{ 2 } { K }_{ 1 }{ x }_{ 1 }^{ 2 } }{ \frac { 1 }{ 2 } { K }_{ 2 }{ x }_{ 2 }^{ 2 } } =\frac { \left( \frac { F }{ { x }_{ 1 } } \right) \times { x }_{ 1 }^{ 2 } }{ \left( \frac { F }{ { x }_{ 2 } } \right) \times { x }_{ 2 }^{ 2 } } =\frac { { x }_{ 1 } }{ { x }_{ 2 } } \)
As x₁ < x₂
\(\therefore\) W₁ < W₂ or W₂ > W₁

(i) There are two external forces on the bob : gravity and the tension (T ) in the string. The latter does no work since the displacement of the bob is always normal to the string. The potential energy of the bob is thus associated with the gravitational force only. The total mechanical energy E of the system is conserved. We take the potential energy of the system to be zero at the lowest point A. Thus, at A :
\(E=\frac{1}{2} m v_{0}^{2}\)   ...(i)
\(T_{A}-m g=\frac{m w_{0}^{2}}{L}\) [Newton’s Second Law]
where T_Ais the tension in the string at A. At the highest point C, the string slackens, as the tension in the string (T_C ) becomes zero.
Thus, at C
\(E=\frac{1}{2} m v_{c}^{2}+2 m g L\)  .....(ii)
\(m g=\frac{m v_{c}^{2}}{L}\) [Newton’s Second Law] .....(iii)
where v_C is the speed at C. From Eqs.(ii) and (iii)
\(E=\frac{5}{2} m g L\)
Equating this to the energy at A
\(\frac{5}{2} m g L=\frac{m}{2} v_{o}^{2}\)
or  \(v_{o}=\sqrt{5 g L}\)
(ii) It is clear from Eq (iii)
\(v_{C}=\sqrt{g L}\)
At B, the energy is
\(E=\frac{1}{2} m v_{B}^{2}+m g L\)
Equating this to the energy at A and employing the result from (i), namely \(v_{o}^{2}=5 g L\)
\(\frac{1}{2} m v_{B}^{2}+m g L=\frac{1}{2} m w_{o}^{2}\)
\(=\frac{5}{2} m g L\)
\(\therefore v_{B}=\sqrt{3 g L}\)
(iii) The ratio of the kinetic energies at B and C is :
\(\frac{K_{B}}{K_{C}}=\frac{\frac{1}{2} m v_{B}^{2}}{\frac{1}{2} m v_{C}^{2}}=\frac{3}{1}\) 
At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution.

(i) W = Fs cos \(\theta\) = 100 x 4 cos 0° = 400 J
(ii) Resistive force opposes the applied force. Box moves at 180^o to the resistive force.
Therefore W = Fs .cos 180 = -400 J. Since motion is along horizontal and gravity is along the vertical, therefore workdone by gravity is
W = Fs cos 90 = zero.
(iii) Force applied = tension in coupling = 2000 N
As P = Fv cos\(\theta\) = 2000 x 20 cos 0 ^o = 40000W = 40 kW
(iv) Work done = Power x time = 40 kW x 60 s = 2400 kJ

A) If both assertion and reason are true and the reason is the correct explanation of the assertion.
The work done, \(W=\vec{F} \cdot \vec{s}=F s \cos \theta\), when a person walk on a horizontal road with load on his head then θ = 90∘. Hence W = Fscos90^∘= 0 Thus no work is done by the person.

(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
In close loop, s = 0, and so W = Fs = 0.

(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.