Work Energy and Power MCQ Questions & Answers in Basic Physics | Physics

Extra power, $$P = mg\sin \theta \times v$$
$$ = Mg \times \frac{h}{s} \times v = \frac{{Mghv}}{s}.$$

$$\eqalign{ & k = 5 \times {10^3}\,N/m \cr & W = \frac{1}{2}k\left( {x_2^2 - x_1^2} \right) \cr & = \frac{1}{2} \times 5 \times {10^3}\left[ {{{\left( {0.1} \right)}^2} - {{\left( {0.5} \right)}^2}} \right] \cr & = \frac{{5000}}{2} \times 0.15 \times 0.05 \cr & = 18.75\,Nm \cr} $$

$$\eqalign{ & \frac{1}{2}M{v^2} = \frac{1}{2}k{L^2}\,\,\,\, \Rightarrow v = \sqrt {\frac{k}{M}} .L \cr & {\text{Momentum}} = M \times v = M \times \sqrt {\frac{k}{M}} .L = \sqrt {kM} .L \cr} $$

Given, $${K_P} > {K_Q}$$
In case $$\left( a \right),$$  the elongation is same
i.e. $${x_1} = {x_2} = x$$
So, $${W_P} = \frac{1}{2}{K_P}{x^2}\,{\text{and}}\,{W_Q} = \frac{1}{2}{K_Q}{x^2}$$
$$\eqalign{ & \therefore \frac{{{W_P}}}{{{W_Q}}} = \frac{{{K_P}}}{{{K_Q}}} > 1 \cr & \Rightarrow {W_P} > {W_Q} \cr} $$
In case $$\left( b \right),$$ the spring force is same
i.e. $${F_1} = {F_2} = F$$
So, $${x_1} = \frac{F}{{{K_P}}},\,{x_2} = \frac{F}{{{K_Q}}}$$
$$\eqalign{ & \therefore {W_P} = \frac{1}{2}{K_P}x_1^2 = \frac{1}{2}{K_P}\frac{{{F^2}}}{{K_P^2}} = \frac{1}{2}\frac{{{F^2}}}{{{K_P}}} \cr & {\text{and}}\,{W_Q} = \frac{1}{2}{K_Q}x_2^2 = \frac{1}{2}{K_Q} \cdot \frac{{{F^2}}}{{K_Q^2}} = \frac{1}{2}\frac{{{F^2}}}{{{K_Q}}} \cr & \therefore \frac{{{W_P}}}{{{W_Q}}} = \frac{{{K_Q}}}{{{K_P}}} < 1 \cr & \Rightarrow {W_P} < {W_Q} \cr} $$

From the law of conservation of momentum
we know that,
$$\eqalign{ & {m_1}{u_1} + {m_2}{u_2} + ..... = {m_1}{v_1} + {m_2}{v_2} + ..... \cr & {\text{Given}}\,\,{m_1} = m,{m_2} = 2m\,{\text{and}}\,{m_3} = 3m \cr & {\text{and}}\,{u_1} = 3u,{u_2} = 2u\,{\text{and}}\,{u_3} = u \cr} $$
Let the velocity when they stick $$ = \vec v$$
Then, according to question,

$$\eqalign{ & m \times 3u\left( {\hat i} \right) + 2\;m \times 2u\left( { - \hat i\cos {{60}^ \circ } - \hat j\sin {{60}^ \circ }} \right) + 3\;m \times u\left( { - \hat i\cos {{60}^ \circ } + \hat j\sin {{60}^ \circ }} \right) = \left( {m + 2\;m + 3\;m} \right)\vec v \cr & \Rightarrow 3mu\hat i - 4mu\frac{{\hat i}}{2} - 4mu\left( {\frac{{\sqrt 3 }}{2}\hat j} \right) - 3mu\frac{{\hat i}}{2} + 3mu\left( {\frac{{\sqrt 3 }}{2}\hat j} \right) = 6m\vec v \cr & \Rightarrow mu\hat i - \frac{3}{2}mu\hat i - \frac{{\sqrt 3 }}{2}mu\hat j = 6m\vec v \cr & \Rightarrow - \frac{1}{2}mu\hat i - \frac{{\sqrt 3 }}{2}mu\hat j = 6m\vec v \cr & \Rightarrow \vec v = \frac{u}{{12}}\left( { - \hat i - \sqrt 3 \hat j} \right) \cr} $$

While moving downhill power
$$\eqalign{ & P = \left( {w\sin \theta + \frac{w}{{20}}} \right)10 \cr & P = \left( {\frac{w}{{10}} + \frac{w}{{20}}} \right)10 = \frac{{3w}}{2} \cr & \frac{P}{2} = \frac{{3w}}{4} = \left( {\frac{w}{{10}} - \frac{w}{{20}}} \right)V \cr & \frac{3}{4} = \frac{v}{{20}} \Rightarrow v = 15\,m/s \cr} $$

$$\therefore $$ Speed of car while moving downhill $$v = 15\,m/s.$$

Kinetic energy acquired by the body $$ = \frac{1}{2}m{v^2}.$$
If body starts fom rest, then final velocity achieved by body in displacement $$d$$ is
$$\eqalign{ & {v^2} = 0 + 2ad = 2ad \cr & {\text{but}}\,a = \frac{F}{m} \cr & \therefore {v^2} = 2\left( {\frac{F}{m}} \right)d \cr} $$
Hence, $$KE = \frac{1}{2}m \times 2\left( {\frac{F}{m}} \right)d = Fd$$
or $$KE$$  acquired = Work done
$$ = F \times d = {\text{constant}}$$
So, $$KE$$  acquired is independent of mass $$m.$$

Let
$$m =$$  mass of boy,
$$M=$$  mass of man
$$v =$$  velocity of boy,
$$V =$$  velocity of man
$$\frac{1}{2}M{V^2} = \frac{1}{2}\left[ {\frac{1}{2}m{v^2}} \right]\,......\left( {\text{i}} \right)$$
$$\frac{1}{2}M{\left( {V + 1} \right)^2} = 1\left[ {\frac{1}{2}m{v^2}} \right]\,......\left( {{\text{ii}}} \right)$$
$${\text{Putting}}\,m = \frac{M}{2}\,{\text{and}}\,{\text{solving}}\,V = \frac{1}{{\sqrt 2 - 1}}$$

Note : In a conservative field work done does not depend on the path. The gravitational field is a conservative field.
$$\therefore \,\,{W_1} = {W_2} = {W_3}$$

$${\text{Initial}}\,KE. = 0,{\text{Initial}}\,P.E. = 0$$
When the rope is just pulled off the table, final $$K.E = \frac{1}{2}\left( {\lambda \ell } \right){v^2},$$    final $$P.E. = \left( {\lambda \ell } \right)\frac{{g\ell }}{2}$$    time taken $$ = t = \frac{\ell }{v}$$
$$\eqalign{ & {\text{Average power}} = \frac{{{\text{net change in energy}}}}{{{\text{time}}}} \cr & = \frac{{\frac{1}{2}\lambda \ell {v^2} + \frac{{\lambda \ell g\ell }}{2}}}{{\frac{\ell }{v}}} = \frac{1}{2}\lambda {v^3} + \frac{{\lambda \ell vg}}{2} \cr} $$