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

Total energy at the time of projection
$$ = \frac{1}{2}m{v^2} = \frac{1}{2} \times 0.1{\left( {20} \right)^2} = 20\,J$$
Half way up, $$P.E.$$  becomes half the $$P.E.$$  at the top i.e. $$P.E. = \frac{{20}}{2} = 10\,J$$
$$\therefore K.E. = 20 - 10 = 10J.$$

Let $${x_A}$$ and $${x_B}$$ be the position of ends $$A$$ and $$B$$ at time $$t$$ from the block, then stretched length of the spring will be
$${\ell _2} = {x_A} - {x_B}$$
and so the stretch
$$\eqalign{ & \Delta \ell = {\ell _2} - {\ell _1} = \left( {{x_A} - {x_B}} \right) - {\ell _1}\left( {{\ell _1}\,{\text{natural}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{spring}}} \right) \cr & {\text{So,}}\,\,U = \frac{1}{2}k\Delta {\ell ^2} = \frac{1}{2}k{\left[ {\left( {{x_A} - {x_B}} \right) - {\ell _1}} \right]^2} \cr & P = \frac{{dU}}{{dt}} = \frac{1}{2}k \cdot 2\left( {{x_A} - {x_B} - {\ell _1}} \right)\left( {\frac{{d{x_A}}}{{dt}} - \frac{{d{x_B}}}{{dt}}} \right) \cr & P = F\left( {{v_A} - {v_B}} \right)\,\,F = \frac{P}{{{v_A} - {v_B}}} \cr & \Delta \ell = \frac{F}{k} = \frac{P}{{\left( {{v_A} - {v_B}} \right)k}} = \frac{{20}}{{\left( {4 - 2} \right) \times 100}} \cr & \Delta \ell = 0.1\,m = 10\,cm \cr} $$

If just after collision, relative velocity $$ = v$$  then
$$\eqalign{ & \frac{v}{u} = \frac{1}{2} \cr & \therefore {\omega _{{\text{rel}}}} = \frac{v}{r} = \frac{u}{{2r}} \cr} $$
$$\therefore $$ time between $${1^{{\text{st}}}}$$ and $${2^{{\text{nd}}}}$$ collision, $$t = \frac{{2\pi }}{{{\omega _{{\text{rel}}}}}}$$
$$ = \frac{{4\pi r}}{u}$$

$$\eqalign{ & \frac{{{\text{Number of particles string the blades}}}}{{{\text{time}}}} \propto {\text{Velocity of wind}} \cr & K.E.{\text{ of particle}} \propto {\left( {{\text{Velocity of wind}}} \right)^2}{\text{ }} \cr & \therefore {\text{Power output}} \propto \left( {\frac{{{\text{No}}{\text{. of particles striking}}}}{{{\text{time}}}}} \right) \times \left( {K.E.{\text{ of particle}}} \right) \cr & \therefore {\text{Power output}} \propto {v^3} \cr} $$

By work-$$KE$$  theorem, we have change in $$KE = $$  work done by all of the forces. Work done by gravitational force,
$${W_g} = mgh = {10^{ - 3}} \times 10 \times 1 \times {10^3} = 10\,J$$
Now, from work-$$KE$$  theorem, we have $$\Delta K = {W_{{\text{gravity}}}} + {W_{{\text{air}}\,{\text{resistance}}}}$$
$$\eqalign{ & \Rightarrow \frac{1}{2} \times m{v^2} = mgh + {W_{{\text{air}}\,{\text{resistance}}}} \cr & \Rightarrow {W_{{\text{air}}\,{\text{resistance}}}} = \frac{1}{2}m{v^2} - mgh \cr & = {10^{ - 3}}\left( {\frac{1}{2} \times 50 \times 50 - 10 \times {{10}^3}} \right) \cr & = - 8.75\,J \cr} $$

Volume of water to raise $$ = 22380\,l$$
$$\eqalign{ & = 22380 \times {10^{ - 3}}\;{m^3} \cr & P = \frac{{mgh}}{t} = \frac{{V\rho gh}}{t} \Rightarrow t = \frac{{V\rho gh}}{P} \cr & t = \frac{{22380 \times {{10}^{ - 3}} \times {{10}^3} \times 10 \times 10}}{{10 \times 746}} = 15\,\min \cr} $$

Work done in moving the object from $$x = 0$$  to $$x = 6\,m,$$  is given by area under the curve.
$$W =$$  Area of square + area of triangle
$$ = 3 \times 2 + \frac{1}{2} \times 3 \times 3 = 9 + 4.5 = 13.5\,J$$

$$\eqalign{ & {\text{Given:}}\,\overrightarrow F = 3\hat i + \widehat j \cr & \overrightarrow {{r_1}} = \left( {2\hat i + \hat k} \right),\overrightarrow {{r_2}} = \left( {4\hat i + 3\hat j - \vec k} \right) \cr & \overrightarrow r = \overrightarrow {{r_2}} - \overrightarrow {{r_1}} = \left( {4\hat i + 3\hat j - \vec k} \right) - \left( {2\hat i + \hat k} \right) \cr & {\text{or}}\,\overrightarrow r = 2\widehat i + 3\widehat j - 2\widehat k \cr} $$
So work done by the given force $$w = \overrightarrow f .\overrightarrow r $$
$$ = \left( {3\hat i + \hat j} \right).\left( {2\hat i + 3\hat j - 2\hat k} \right) = 6 + 3 = 9\,J$$

Let $$x$$ be the extension in the spring.
Applying conservation of energy
$$\eqalign{ & Mgx - \frac{1}{2}k{x^2} = 0 - 0 \cr & \Rightarrow x = \frac{{2Mg}}{k} \cr} $$

Given, $$K{E_1} = K{E_2}$$
$$\eqalign{ & \frac{1}{2}{m_1}v_1^2 = \frac{1}{2}{m_2}v_2^2 \cr & {\text{or}}\,\,\frac{{v_2^2}}{{v_1^2}} = \frac{{{m_1}}}{{{m_2}}}\,\,{\text{or}}\,\,\frac{{{v_2}}}{{{v_1}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}} \cr & {\text{As,}}\,\,{p_2} = {m_2}{v_2}{\text{ and }}{p_1} = {m_1}{v_1} \cr & \therefore \frac{{{p_2}}}{{{p_1}}} = \frac{{{m_2}{v_2}}}{{{m_1}{v_1}}} = \frac{{{m_2}}}{{{m_1}}}\sqrt {\frac{{{m_1}}}{{{m_2}}}} = \sqrt {\frac{{m_2^2{m_1}}}{{m_1^2{m_2}}}} \cr & \frac{{{p_2}}}{{{p_1}}} = \sqrt {\frac{{{m_2}}}{{{m_1}}}} \cr & {\text{or}}\,\,\frac{{{p_1}}}{{{p_2}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}} \cr & {\text{Here,}}\,\,{m_1} = 1\;g,{m_2} = 9\;g \cr & \therefore \frac{{{p_1}}}{{{p_2}}} = \sqrt {\frac{1}{9}} = \frac{1}{3} \cr} $$
Alternative
The relation between $$KE$$  and $$p$$ is given by
$$\eqalign{ & KE = \frac{{{p^2}}}{{2m}} \cr & \Rightarrow {p^2} = 2mKE \cr & \Rightarrow p = \sqrt {2mKE} \cr} $$
If $$KE$$  of two bodies are equal.
$$\eqalign{ & {\text{So,}}\,\,{p_1} \propto \sqrt {{m_1}} \cr & {\text{and}}\,\,{p_2} \propto \sqrt {{m_2}} \cr & \Rightarrow \frac{{{p_1}}}{{{p_2}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}} \cr & = \sqrt {\frac{1}{9}} = \frac{1}{3} \cr} $$