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

Given, Velocity of water $$v = 2\,m/s$$
Mass per unit length of water in the pipe $$ = 100\,kg/m$$
So, $${\text{power}} = \left( {{\text{mass per unit length of water in pipe}}} \right) \times {v^3}$$
$$\eqalign{ & = \frac{m}{l} \times {v^3} \cr & = 100 \times 2 \times 2 \times 2 \cr & = 800\,W \cr} $$

$${\text{Power}} = \frac{{{\text{ work done}}}}{{{\text{ time}}}}$$
Therefore power of $$A,$$ $${P_A} = \frac{{mgh}}{{{t_A}}}$$
and power of $$B,$$ $${P_B} = \frac{{mgh}}{{{t_B}}}$$
$$\therefore \frac{{{P_A}}}{{{P_B}}} = \frac{{{t_B}}}{{{t_A}}} = \frac{4}{2} = 2:1$$

From work-energy theorem,
Work done = Change in $$KE$$
$$\eqalign{ & \Rightarrow W = {K_f} - {K_i} \cr & \Rightarrow {K_f} = W + {K_i} = \int_{{x_1}}^{{x_2}} {Fxdx + \frac{1}{2}m{v^2}} \cr & = \int_{20}^{30} { - 0.1x\,dx + \frac{1}{2} \times 10 \times {{10}^2}} \cr} $$
$$\eqalign{ & = - 0.1\left[ {\frac{{{x^2}}}{2}} \right]_{20}^{30} + 500 \cr & = - 0.05\left[ {{{30}^2} - {{20}^2}} \right] + 500 \cr & = - 0.05\left[ {900 - 400} \right] + 500 \cr & \Rightarrow {K_f} = - 25 + 500 = 475\,J \cr} $$

From energy conservation $$\frac{1}{2}mv_{\max }^2 = mg\left( {{H_2} - {H_1}} \right)$$
Here, $${{H_1}} =$$  minimum height of swing from the earth's surface
$$ = 0.75\,m$$
$${{H_2}} =$$  maximum height of swing from earth's surface
$$ = 2\,m$$
$$\eqalign{ & \therefore \frac{1}{2}mv_{\max }^2 = mg\left( {2 - 0.75} \right) \cr & {\text{or}}\,\,{v_{\max }} = \sqrt {2 \times 10 \times 125} \cr & = \sqrt {25} \cr & = 5\,m/s \cr} $$

For any uniform rod, the mass is supposed to be concentrated at its centre.

$$\therefore $$ height of the mass from ground is, $$h = \left( {\frac{l}{2}} \right)\sin {30^ \circ }$$
$$\therefore $$ Potential energy of the rod
$$\eqalign{ & = m \times g \times \frac{\ell }{2}\sin {30^ \circ } \cr & = m \times g \times \frac{\ell }{2} \times \frac{1}{2} = \frac{{mg\,\ell }}{4} \cr} $$

$$\eqalign{ & K.E. = \frac{1}{2}m{v^2} = \frac{1}{2} \times 0.1 \times 25 = 1.25\,J \cr & W = - mgh = - \left( {\frac{1}{2}m{v^2}} \right) = - 1.25\,J \cr & \left[ {\because mgh = \frac{1}{2}m{v^2}{\text{ by energy conservation}}} \right] \cr} $$

$$\eqalign{ & W = F\,s\,\cos \theta ,\cos \theta = \frac{W}{{Fs}} = \frac{{25}}{{5 \times 10}} = \frac{1}{2}, \cr & \theta = {60^ \circ }. \cr} $$

Work done = area under $$F-x$$  graph
$$\eqalign{ & = {\text{area of trapezium }}OABC = \frac{1}{2}\left( {3 + 6} \right)\left( 3 \right) \cr & = 13.5\,J \cr} $$

Initial momentum = Final momentum
$$\therefore {m_1}{v_1} + {m_2}{v_2} = \left( {{m_1} + {m_2}} \right)v$$
Given, $${v_1} = 36\,km/h = 36 \times \frac{5}{{18}} = 10\,m/s,$$
$$\eqalign{ & {v_2} = 0 \cr & {m_1} = 2\,kg,{m_2} = 3\,kg \cr & \therefore v = \frac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}} \cr & = \frac{{2 \times 10 + 3 \times 0}}{{2 + 3}} \cr & {\text{or}}\,\,v = \frac{{20}}{5} = 4\;\,m/s \cr} $$
Loss in kinetic energy
$$\eqalign{ & = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 - \frac{1}{2}\left( {{m_1} + {m_2}} \right){v^2} \cr & = \frac{1}{2} \times 2 \times {\left( {10} \right)^2} + 0 - \frac{1}{2}\left( {2 + 3} \right) \times {\left( 4 \right)^2} \cr & = 100 - 40 = 60\,J \cr} $$

$$u = 0;\,\,\,\,v = u + aT;\,\,\,\,v = aT$$
Instantaneous power $$ = F \times v = m\,.\,a\,.\,aT = m\,.\,{a^2}.\,\,t$$
$$\therefore $$ Instantaneous power $$ = m\frac{{{v^2}}}{{{T^2}}}t$$