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

$$\eqalign{ & m = 1\,kg,a = 1\,m/{s^2},\mu = 0.1 \cr & f = \mu mg = 0.1 \times 1 \times 10 = 1N \cr & {F_{{\text{ext}}}} = ma + f \cr & {F_{{\text{ext}}}} = 1 + 1 = 2N \cr & v = u + at = 0 + 1 \times 2 = 2\,m/\sec . \cr & \therefore P = {F_{{\text{ext}}}} \cdot v = 2 \times 2 = 4\,watt. \cr} $$

For a given spring, $$u = \frac{1}{2}k{x^2}$$
$$\therefore \frac{{{u_2}}}{{{u_1}}} = \frac{{\frac{1}{2}kx_2^2}}{{\frac{1}{2}kx_1^2}} = \frac{{{{\left( {3x} \right)}^2}}}{{{x^2}}} = 9:1$$

$$\eqalign{ & {\text{As}}\,{u_2} = 0\,{\text{and}}\,{m_1} = {m_2},{\text{therefore}}\,{\text{from}} \cr & {m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2} \cr & {\text{we}}\,{\text{get}}\,{u_1} = {v_1} + {v_2} \cr & {\text{Also,}}\,e = \frac{{{v_2} - {v_1}}}{{{u_1}}} = \frac{{{v_2} - {v_1}}}{{{v_2} + {v_1}}} = \frac{{\frac{{1 - {v_1}}}{{{v_2}}}}}{{\frac{{1 + {v_1}}}{{{v_2}}}}}, \cr & {\text{which}}\,{\text{gives}}\,\frac{{{v_1}}}{{{v_2}}} = \frac{{1 - e}}{{1 + e}} \cr} $$

The average speed of the athlete
$$v = \frac{{100}}{{10}} = 10\,m/s\,\,\,\,\,\,\,\,\,\therefore K.E. = \frac{1}{2}m{v^2}$$
If mass is 40 kg then, K.E. $$ = \frac{1}{2} \times 40 \times {\left( {10} \right)^2} = 2000\,J$$
If mass is 100 kg then, K.E. $$ = \frac{1}{2} \times 100 \times {\left( {10} \right)^2} = 5000\,J$$

When force is same
$$\eqalign{ & W = \frac{1}{2}k{x^2} \cr & W = \frac{1}{2}k\frac{{{F^2}}}{{{k^2}}}\,\,\left[ {\because \,F = kx} \right] \cr & \therefore W = \frac{{{F^2}}}{{2x}} \cr & {\text{As}}\,\,{W_1} > {W_2} \cr & \therefore {k_1} < {k_2} \cr} $$
When extension is same
$$\eqalign{ & W \propto k\,\,\left( {\because x{\text{ is same}}} \right) \cr & \therefore {W_1} < {W_2} \cr} $$
Statement I is false and statement 2 is true.

$$\eqalign{ & y = 10 - \sqrt {{{10}^2} - {4^2}} = 0.84\,m \cr & {\text{Thus}}\,\,\frac{1}{2}m{v^2} = mgy \cr & {\text{or}}\,\,v = \sqrt {2gy} = \sqrt {2 \times 9.8 \times 0.84} \cr & = 4.05\,m/s. \cr} $$

Using relative velocity, time of flight before collision will be
$$t = \frac{{20}}{{20}} = 1$$

By $$COM$$  at the time of collision
$$\eqalign{ & 3 \times 10 - 1 \times 10 = 4 \times v \cr & 2 \times 10 = 4 \times v;5 = v \cr & v = 5\,m/s \cr} $$
For $$1$$ - $$D$$ motion
$$\eqalign{ & {v^2} = {u^2} + 2as = {5^2} + 2 \times 10 \times 15 = 25 + 300 = 325 \cr & K = 650\,J \cr} $$

For the smaller block to move $$k{x_0} = \mu mg$$   and from work energy theorem
$$\eqalign{ & - \mu Mg{x_0} - \frac{1}{2}kx_0^2 = - \frac{1}{2}Mv_0^2 \cr & + \mu Mg\left( {\frac{{\mu mg}}{k}} \right) + \frac{1}{2}k{\left( {\frac{{\mu mg}}{k}} \right)^2} = \frac{1}{2}M{v^2} \cr & v = \mu g\sqrt {\frac{{\left( {2M + m} \right)m}}{{kM}}} \cr} $$

$$\eqalign{ & {W_{{\text{gravity }}}} + {W_F} = \Delta K - 10 \times g\left( {10 - 10\cos {{60}^ \circ }} \right) + 200 \times 10\sin {60^ \circ } \cr & = \frac{1}{2} \times 10 \times v_f^2 - 0 \cr & \therefore {v_f} = 17\,m/s \cr} $$

$$\eqalign{ & {\text{Power}}\,\vec F \cdot \vec V = PA\vec V = \rho ghAV\,\,\left[ {\because P = \frac{F}{A}\,{\text{and}}\,P = \rho gh} \right] \cr & = \frac{{13.6 \times {{10}^3} \times 10 \times 150 \times {{10}^{ - 3}} \times 0.5 \times {{10}^{ - 3}}}}{{60}} \cr & = \frac{{102}}{{60}} = 1.70\,watt \cr} $$