$${\text{Mechanical efficiency}} = \frac{{{\text{Output work }}}}{{{\text{Input work}}}}$$
The output work will increase because the friction becomes less. Thus the mechanical efficiency increases.

$$\eqalign{ & x = 3t - 4{t^2} + {t^3} \cr & \frac{{dx}}{{dt}} = 3 - 8t + 3{t^2} \cr} $$
Acceleration $$ = \frac{{{d^2}x}}{{d{t^2}}} = - 8 + 6t$$
Acceleration after $$4\sec = - 8 + 6 \times 4 = 16\,m{s^{ - 2}}$$
Displacement in $$4\sec = 3 \times 4 - 4 \times {4^2} + {4^3} = 12\,m$$
$$\eqalign{ & \therefore {\text{Work}} = {\text{Force}} \times {\text{displacement}} \cr & = {\text{Mass}} \times {\text{acc}}{\text{.}} \times {\text{disp}}{\text{.}} = 3 \times {10^{ - 3}} \times 16 \times 12 = 576\,mJ \cr} $$

Given, $${u_1} = 3m/s,\,{u_2} = - 5\,m/s,\,{m_1} = {m_2} = m$$
According to principle of conservation of linear momentum,
$$\eqalign{ & {m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2} \cr & m \times 3 - m \times 5 = m{v_1} + m{v_2} \cr} $$
$${\text{or}}\,\,{v_1} + {v_2} = - 2\,.....\left( {\text{i}} \right)$$
In an elastic collision, $$e = \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}$$
$$\eqalign{ & \Rightarrow {v_2} - {v_1} = e\left( {{u_1} - {u_2}} \right) \cr & \Rightarrow {v_2} - {v_1} = \left( 1 \right)\left( {3 + 5} \right)\,\,\left( {\because e = 1} \right) \cr & \Rightarrow {v_1} - {v_2} = - 8\,......\left( {\text{ii}} \right) \cr} $$
Adding Eqs. (i) and (ii), we obtain
$$\eqalign{ & 2{v_1} = - 10 \cr & \Rightarrow {v_1} = - 5\,m/s \cr} $$
From Eq. (i), $${v_2} = - 2 - {v_1} = - 2 + 5 = 3\,m/s$$
Thus, $${v_1} = - 5\,m/s,\,{v_2} = + 3\,m/s$$
If two bodies collide elastically, then their velocities are interchanged. Since, it is an elastic collision hence, velocities after collision will be $$-5\,m/s$$  and $$3\,m/s.$$

$$\eqalign{ & W = mgh + \frac{1}{2}k{x^2} = mg\left( {a\sin \theta } \right) + \frac{1}{2}k{\left( {a\theta } \right)^2} \cr & = W\,a\sin \theta + \frac{1}{2}k{a^2}{\theta ^2}. \cr} $$

$$\eqalign{ & {\text{Power}} = \frac{{{\text{Work}}\,{\text{done}}}}{{{\text{Time}}}} = \frac{{\frac{1}{2}m\left( {{v^2} - {u^2}} \right)}}{t} \cr & P = \frac{1}{2} \times \frac{{2.05 \times {{10}^6} \times \left[ {{{\left( {25} \right)}^2} - {{\left( 5 \right)}^2}} \right]}}{{5 \times 60}} \cr & P = 2.05 \times {10^6}W = 2.05\,MW \cr} $$

Let $$x$$ be the distance travelled by the particle along the $${37^ \circ }$$ incline, after which it comes to rest. From work-energy theorem, Work done by gravity on block on incline $${53^ \circ } + $$  work done by friction on block on incline $${53^ \circ } + $$  work done by friction on ground level $$+$$ work done by gravity on block on incline $${37^ \circ } + $$  work done by friction on block on incline $${37^ \circ } = $$  change in $$K.E.$$  $$mg\left( 8 \right) - \mu mg\cos {53^ \circ }\left( {10} \right) - \mu mg\left( {10} \right) - \mu mg\cos {37^ \circ }\left( x \right) - mg\sin {37^ \circ }\left( x \right) = 0 - \frac{1}{2}\left( 1 \right){\left( {10} \right)^2}$$
On solving we get $$x = 1\,m$$

As one piece retraces its path, the speed of this piece just after explosion should be $$v\cos \theta $$

Applying conservation of linear momentum at the highest point;
$$\eqalign{ & m\left( {v\cos \theta } \right) = \frac{m}{2} \times v' - \frac{m}{2} \times v\cos \theta \cr & 3v\cos \theta = v' \cr} $$

KEY CONCEPT
The force constant of a spring is inversely proportional to the length of the spring.
Let the original length of spring be $$L$$  and spring constant is $$k$$  (given)
Therefore,
$$\eqalign{ & k \times L = \frac{{2L}}{3} \times k' \cr & \Rightarrow k' = \frac{3}{2}k \cr} $$

$$\eqalign{ & P = \frac{W}{t}.\,{\text{Here,}}\,P = 2\;kW = 2000\;W. \cr & W = Mgh = M \times 10 \times 10 = 100M \cr & {\text{and}}\,\,t = 60\,s. \cr} $$
This gives, $$M = 1200\,kg$$
Its volume = 1200 litre as 1 litre of water contains $$1\,kg$$  of its mass.

$$\eqalign{ & {P_{{\text{out}}}} = \frac{{mgh}}{t} = \frac{{9000 \times 10 \times 10}}{{5 \times 60}} = 3000\,W \cr & {P_{{\text{in}}}} = 10 \times {10^3}W. \cr & \therefore \eta = \frac{{{P_{{\text{out}}}}}}{{{P_{{\text{in}}}}}} \times 100 = \frac{{3000}}{{10 \times {{10}^3}}} \times 100 = 30\% \cr} $$