Given, pressure $$= 150\,mm$$   of $$Hg$$
Pumping rate of heart of a man
$$ = \frac{{dV}}{{dt}} = \frac{{5 \times {{10}^{ - 3}}}}{{60}}\,{m^3}/s$$
Power of heart $$ = P \cdot \frac{{dV}}{{dt}} = \rho gh \cdot \frac{{dV}}{{dt}}\,\,\left[ {p = \rho gh} \right]$$
$$\frac{{\left( {13.6 \times {{10}^3}\,kg/{m^3}} \right)\left( {10 \times 0.15 \times 5 \times {{10}^{ - 3}}} \right)}}{{60}} = 1.70\,W$$

Mass of over hanging chain $$m' = \frac{4}{2} \times \left( {0.6} \right)kg$$
Let at the surface $$PE =0$$
C.M. of hanging part $$= 0.3 \,m$$   below the table
$${U_i} = - m'gx = - \frac{4}{2} \times 0.6 \times 10 \times 0.30$$
$$\Delta U = m'gx = 3.6\,J = $$     Work done in putting the entire chain on the table.

According to question, a body of mass $$1\,kg$$  begins to move under the action of time dependent force,
$$F = \left( {2t\hat i + 3{t^2}\hat j} \right)N$$
where $${\hat i}$$ and $${\hat j}$$ are unit vectors along $$X$$ and $$Y$$-axes.
$$\eqalign{ & \because F = ma \cr & \Rightarrow a = \frac{F}{m} \cr & \Rightarrow a = \frac{{\left( {2t\hat i + 3{t^2}\hat j} \right)}}{1}\,\,\,\left( {\because m = 1\,kg} \right) \cr & \Rightarrow a = \left( {2t\hat i + 3{t^2}\hat j} \right)m/{s^2} \cr & \because {\text{acceleration,}}\,\,a = \frac{{dv}}{{dt}} \cr & \Rightarrow dv = adt\,......\left( {\text{i}} \right) \cr} $$
Integrating both sides, we get
$$\eqalign{ & \int {dv} = \int a \,dt = \int {\left( {2t\hat i + 3{t^2}\hat j} \right)dt} \cr & v = {t^2}\hat i + {t^3}\hat j \cr} $$
$$\because $$ Power developed by the force at the time $$t$$ will be given as
$$\eqalign{ & P = F \cdot v = \left( {2t\hat i + 3{t^2}\hat j} \right).\left( {{t^2}\hat i + {t^3}\hat j} \right) \cr & = \left( {2t.{t^2} + 3{t^2}.{t^3}} \right) \cr & P = \left( {2{t^3} + 3{t^5}} \right)W \cr} $$

According to law of conservation of kinetic energy,
we have $$\frac{1}{2}M{v^2} + 0 = \frac{1}{2}M{\left( {\frac{v}{3}} \right)^2} + \frac{1}{2}Mv_2^2$$
$$\eqalign{ & \Rightarrow {v^2} = \frac{{{v^2}}}{9} + v_2^2 \cr & \Rightarrow {v^2} - \frac{{{v^2}}}{9} = v_2^2 \Rightarrow \frac{{8{v^2}}}{9} \cr} $$
Velocity of second block after collision $${v_2} = \frac{{2\sqrt 2 }}{3}v$$

In the frame of lift displacement is zero,
$${\text{so}}\,\,W = F.S = T.S = 0.$$

By using mechanical energy conservation $$v = \frac{{10}}{3} = 3.33\,m/s$$

Let $$v$$ be velocity of combined mass after collision.
Applying law of conservation of linear momentum
Initial momentum = Final momentum
$$\eqalign{ & m \times 3 + 2m \times 0 = \left( {m + 2m} \right)v \cr & {\text{or}}\,\,3m = 3mv\,\,{\text{or}}\,\,v = 1\,km/h \cr} $$

Given, $$x = 3t - 4{t^2} + {t^3}$$
So, velocity $$v = \frac{{dx}}{{dt}} = 3 - 8t + 3{t^2}$$
$$\eqalign{ & {\text{At}}\,\,t = 0\;s,\,\,{v_1} = 3 - 0 + 0 = 3\,m/s \cr & {\text{At}}\,\,t = 4\,s,\,\,{v_2} = 3 - 8 \times 4 + 3 \times {4^2} \cr & = 3 - 32 + 48 = 19\,m/s \cr} $$
Now, work done during $$t = 0$$  to $$t = 4\,s$$
= gain in kinetic energy
$$\eqalign{ & = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = \frac{1}{2}m\left( {v_2^2 - v_1^2} \right) \cr & = \frac{1}{2} \times 3 \times {10^{ - 3}}\left[ {{{\left( {19} \right)}^2} - {{\left( 3 \right)}^2}} \right]\,\,\,\left[ {{\text{Using, }}{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right] \cr & = 1.5 \times {10^{ - 3}} \times \left[ {\left( {19 + 3} \right)\left( {19 - 3} \right)} \right] \cr & = 1.5 \times {10^{ - 3}} \times 22 \times 16 \cr & = 528 \times {10^{ - 3}}\,J \cr & = 528\,mJ \cr} $$

Suppose a ball rebounds with speed $$v,$$
$$\eqalign{ & v = \sqrt {2gh} = \sqrt {2 \times 10 \times 20} \cr & = 20\,m/s \cr} $$
Energy of a ball just after rebound,
$$E = \frac{1}{2}m{v^2} = 200\,m$$
As, $$50\% $$ of energy loses in collision means just before collision energy is $$400\,m.$$
According to law of conservation of energy, we have
$$\eqalign{ & \frac{1}{2}mv_0^2 + mgh = 400\,m \cr & \Rightarrow \frac{1}{2}mv_0^2 + m \times 10 \times 20 = 400\,m \cr & \Rightarrow {v_0} = 20\,m/s \cr} $$

Loss in gravitational $$P.E. = $$   gain in spring $$P.E.$$
$$\eqalign{ & mgh = \frac{1}{2}k{\left( {h\cot \alpha - h} \right)^2}\,\,{\text{or}}\,\,\left( {\cot \alpha - 1} \right) = \sqrt {\frac{{2mg}}{{kh}}} \cr & {\text{or}}\,\,\cot \alpha = 1 + \sqrt {\frac{{2mg}}{{kh}}} \cr} $$