$$\eqalign{ & E = \frac{{{P^2}}}{{2m}} \cr & \therefore E \propto \frac{1}{m}\,\,\left[ {{\text{If}}\,P = {\text{constant}}} \right] \cr} $$
i.e. the lightest particle will possess maximum kinetic energy and in the given a option mass of electron is minimum.

Work done by the force $$ = {\text{force}} \times {\text{displacement}}$$
$${\text{or}}\,\,W = F \times s\,......\left( {\text{i}} \right)$$
But from Newton's 2^nd law, we have
$${\text{Force}} = {\text{mass}} \times {\text{acceleration}}$$
$${\text{i}}{\text{.e}}{\text{.}}\,\,F = ma\,......\left( {{\text{ii}}} \right)$$
Hence, from Eqs. (i) and (ii), we get
$$W = mas = m\left( {\frac{{{d^2}s}}{{d{t^2}}}} \right)s\left( {\because a = \frac{{{d^2}s}}{{d{t^2}}}} \right)\,\,......\left( {{\text{iii}}} \right)$$
Now, we have, $$s = \frac{1}{3}{t^2}$$
$$\eqalign{ & \therefore \frac{{{d^2}s}}{{d{t^2}}} = \frac{d}{{dt}}\left[ {\frac{d}{{dt}}\left( {\frac{1}{3}{t^2}} \right)} \right] \cr & = \frac{d}{{dt}} \times \left( {\frac{2}{3}t} \right) = \frac{2}{3}\frac{{dt}}{{dt}} = \frac{2}{3} \cr} $$
Hence, Eq. (iii) becomes
$$W = \frac{2}{3}ms = \frac{2}{3}m \times \frac{1}{3}{t^2} = \frac{2}{9}m{t^2}$$
We have, $$m = 3\,kg,t = 2\,s$$
$$\therefore W = \frac{2}{9} \times 3 \times {\left( 2 \right)^2} = \frac{8}{3}J$$

We know that $$F×v= Power$$
$$\therefore $$ $$F×v=c$$   where $$c=constant$$
$$\eqalign{ & \therefore m\frac{{dv}}{{dt}} \times v = c\,\,\,\,\,\left( {\therefore F = ma = \frac{{mdv}}{{dt}}} \right) \cr & \therefore m\int\limits_0^v {vdv} = c\int\limits_0^t {dt} \cr & \therefore \frac{1}{2}m{v^2} = ct \cr & \therefore v = \sqrt {\frac{{2c}}{m}} \times {t^{\frac{1}{2}}} \cr & \therefore \frac{{dx}}{{dt}} = \sqrt {\frac{{2c}}{m}} \times {t^{\frac{1}{2}}}\,\,\,\,{\text{where }}v = \frac{{dx}}{{dt}} \cr & \therefore \int\limits_0^x {dx} = \sqrt {\frac{{2c}}{m}} \times \int\limits_0^t {{t^{\frac{1}{2}}}} dt \cr & x = \sqrt {\frac{{2c}}{m}} \times \frac{{2{t^{\frac{3}{2}}}}}{3} \cr & \Rightarrow x \propto {t^{\frac{3}{2}}} \cr} $$

Constant power of car $${P_0} = F.V = ma.v$$
$$\eqalign{ & {P_0} = m\frac{{dv}}{{dt}} \cdot v \cr & {P_0}dt = mvdv\,{\text{Integrating}} \cr & {P_0}.t = \frac{{m{v^2}}}{2}\,\,v = \sqrt {\frac{{2{P_0}t}}{m}} \cr & \because {P_0},m\,{\text{and}}\,2\,{\text{are}}\,{\text{constant}} \cr & \therefore v \propto \sqrt t \cr} $$

Work done in displacing the particle,
$$\eqalign{ & W = \vec F.\,\vec x = \left( {5\vec i + 3\vec j + 2\vec k} \right).\left( {2\vec i - \vec j} \right) \cr & = 10 - 3 = 7\,joules \cr} $$

$${m_1}v = - {m_1}{v_1} + {m_2}{v_2},$$
$$e = \frac{{2{v_1}}}{v}$$   for limiting condition
$$\frac{{{m_1}}}{{{m_2}}} < \frac{e}{{2 + e}}$$

Area under graph increases. Hence work done upon the particle from $$A$$ to $$B$$ increases.

Work done is equal to area under the curve in $$F$$-$$d$$ graph.
Work done = Area under $$\left( {F{\text{-}}d} \right)$$  graph $$ = 2 \times \left( {7 - 3} \right) + \frac{1}{2} \times 2 \times \left( {12 - 7} \right)$$
$$\eqalign{ & = 8 + \frac{1}{2} \times 10 \cr & = 8 + 5 = 13\,J \cr} $$

$$\eqalign{ & Pt = nmgh \cr & 2400\left( {60} \right) = n\left( {65} \right)\left( {9.8} \right)\left( 6 \right);n = 37 \cr} $$

$${v^2} = {u^2} + 2gh = {\left( {10} \right)^2} + 2 \times 10 \times 19.5 = 490$$
$$K.E.$$  at the ground $$ = \frac{1}{2}m{v^2} = \frac{1}{2} \times \frac{5}{{1000}} \times 490 = \frac{{49}}{{40}}\,J$$
$$P.E. = mgh = \frac{5}{{1000}} \times 10 \times \left( {\frac{{ - 50}}{{100}}} \right) = - \frac{1}{{40}}\,J$$
∴ Change in energy $$ = \frac{{49}}{{40}} - \left( { - \frac{1}{{40}}} \right)$$
$$ = \frac{{50}}{{40}} = 1.25\,J$$