$$\eqalign{ & K.E. = \frac{1}{2}m{v^2} = \frac{1}{2}m{\left[ {u + at} \right]^2} = \frac{1}{2}m{\left[ {0 + gt} \right]^2} \cr & \therefore K.E. = \frac{1}{2}mg{t^2} \cr & \therefore K.E. \propto {t^2}.....(1) \cr} $$
First the kinetic energy will increase as per equation (1). As the balls touches the ground it starts deforming and loses its K.E. (K.E. converting into elastic potential energy). When the deformation is maximum, K.E. = 0. The ball then again regain its shape when its elastic potential energy changes into K.E. As the ball moves up it loses K.E. and gain gravitational potential energy.
These characteristics are according to graph (B).

$$\eqalign{ & {U_1} = \int\limits_0^{\frac{\ell }{3}} { - \frac{m}{\ell }} gxdx = - \frac{1}{{18}}mg\ell ; \cr & {U_2} = \int\limits_0^\ell { - \frac{m}{\ell }} gxdx = - \frac{1}{2}mg\ell \cr & {\text{loss}}\,{\text{in}}\,P.E. = {U_1} - {U_2} = \frac{4}{9}mgl \cr & = \frac{4}{9} \times 0.1 \times 10 \times 2 = \frac{8}{9}J = {\text{Final}}\,K.E. \cr} $$

$$\eqalign{ & n = \frac{W}{{{\text{input}}}} = \frac{{mgh \times 1000}}{{{\text{input}}}} = \frac{{10 \times 9.8 \times 1 \times 1000}}{{{\text{input}}}} \cr & {\text{Input}} = \frac{{98000}}{{0.2}} = 49 \times {10^4}\,J \cr & {\text{Fat used}} = \frac{{49 \times {{10}^4}}}{{3.8 \times {{10}^7}}} = 12.89 \times {10^{ - 3}}\,kg \cr} $$

Given: Mass of particle, $$M = 10\,g = \frac{{10}}{{1000}}kg$$
radius of circle $$R = 6.4\,cm$$
Kinetic energy $$E$$ of particle $$ = 8 \times {10^{ - 4}}J$$
acceleration $${a_t} = ?$$
$$\eqalign{ & \frac{1}{2}m{v^2} = E \Rightarrow \frac{1}{2}\left( {\frac{{10}}{{1000}}} \right){v^2} = 8 \times {10^{ - 4}} \cr & \Rightarrow {v^2} = 16 \times {10^{ - 2}} \Rightarrow v = 4 \times {10^{ - 1}} = 0.4\,m/s \cr} $$
Now, using
$$\eqalign{ & {v^2} = {u^2} + 2{a_t}s\,\,\left( {s = 4\pi R} \right) \cr & {\left( {0.4} \right)^2} = {0^2} + 2{a_t}\left( {4 \times \frac{{22}}{7} \times \frac{{6.4}}{{100}}} \right) \cr & \Rightarrow {a_t} = {\left( {0.4} \right)^2} \times \frac{{7 \times 100}}{{8 \times 22 \times 6.4}} = 0.1\,m/{s^2} \cr} $$

Power, $$P = F.v = m\frac{{dv}}{{dt}}.v$$
As $$P$$ is constant, $$\frac{v}{{dt}} = {\text{constant}}$$
$$ \Rightarrow v \propto dt \Rightarrow v \propto t$$

Net displacement is $$\left( {3\hat i + 3\hat j} \right)$$
so work done $$ = \overrightarrow F \cdot \Delta \overrightarrow s = 11.25\left( {\widehat i + \widehat j} \right).\left( {3\widehat i + 3\widehat j} \right)$$
$$ = 33.75 \times 2 = 67.50\,J$$

The velocity of particle after falling through height $$h$$
$$u = \sqrt {2gh} \,......\left( {\text{i}} \right)$$
After first rebounding, the velocity of ball is $$eu$$ and after attaining maximum height it will come to the ground with same velocity $$eu.$$ So, after second rebounding its velocity will be $${e^2}u.$$  Similarly, after third fourth ... etc reboundings its velocities will be $${e^2}u,{e^4}u,....$$   etc.
Since, it first rebounds with velocity $$eu$$ so if it attains height $$h$$ then from
$$\eqalign{ & {v^2} = {u^2} - 2gh \cr & \therefore 0 = {e^2}{u^2} - 2g{h_1} \cr & {\text{or}}\,\,{h_1} = \frac{{{e^2}{u^2}}}{{2g}} = \frac{{{e^2}2gh}}{{2g}} = {e^2}h\,\,\left[ {{\text{from}}\,{\text{Eq}}{\text{.}}\left( {\text{i}} \right)} \right] \cr} $$
The same height the ball travels while approaching ground. Now, it rebounds with velocity $${e^2}u$$  so if it attains a height $${h_2}$$ then
$$\eqalign{ & 0 = {e^4}{u^2} - 2g{h_2} \cr & {\text{or}}\,\,{h_2} = {e^4}h \cr} $$
The similar process will follow for further reboundings
Hence, the total distance travelled by the practice before it stops rebounding.
$$\eqalign{ & = h + 2{h_1} + 2{h_2} + ...\infty = h + 2{e^2}h + 2{e^4}h + ...\infty \cr & = h + 2{e^2}h\left( {1 + {e^2} + {e^4} + ...\infty } \right) = h + 2{e^2}h\left( {\frac{1}{{1 - {e^2}}}} \right) \cr & = h\left( {1 + \frac{{2{e^2}}}{{1 - e}}} \right) \cr & = \left( {\frac{{1 + {e^2}}}{{1 - {e^2}}}} \right)h \cr} $$

As the cord is trying to hold the motion of the block, work done by the cord is negative.
$$\eqalign{ & W = - M\left( {g - a} \right)d \cr & = - M\left( {g - \frac{g}{4}} \right)d = \frac{{ - 3Mgd}}{4} \cr} $$

Work done by such force is always zero since force is acting in a direction perpendicular to velocity.
$$\therefore $$ from work-energy theorem $$ = \Delta K = 0$$
$$K$$  remains constant.

Power exerted by a force is given by $$P = F.v$$
When the body is just above the earth’s surface, its velocity is greatest. At this instant, gravitational force is also maximum. Hence, the power exerted by the gravitational force is greatest at the instant just before the body hits the earth.