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

$$\eqalign{ & {x_i} = \frac{{50}}{{500}} = 0.1\,m,{\ell _i} = 0.4 - 0.1 = 0.3\,m \cr & {x_f} = \sqrt {{{0.4}^2} + {{0.3}^2}} - 0.3 = 0.2\;m \cr & 2 \times \frac{1}{2}kx_i^2 + mg \times 0.3 = 2 \times \frac{1}{2}kx_f^2 + \frac{1}{2}m{v^2} \Rightarrow v = 0 \cr} $$

Kinetic energy of a system of particle is zero only when the speed of each particles is zero. And if speed of each particle is zero, the linear momentum of the system of particle has to be zero.
Also the linear momentum of the system may be zero even when the particles are moving. This is because linear momentum is a vector quantity. In this case the kinetic energy of the system of particles will not be zero.
$$\therefore $$ A does not imply B but B implies A.

Initial momentum $$\left( {{p_1}} \right) = p;$$   Final momentum $$\left( {{p_2}} \right) = 1.5p$$   and initial kinetic energy $$\left( {{K_1}} \right) = K.$$
Kinetic energy $$\left( K \right) = \frac{{{p^2}}}{{2m}} \propto {p^2}$$
$$\eqalign{ & {\text{or,}}\,\,\,\frac{{{K_1}}}{{{K_2}}} = {\left( {\frac{{{p_1}}}{{{p_2}}}} \right)^2} = {\left( {\frac{p}{{1.5p}}} \right)^2} = \frac{1}{{2.25}} \cr & {\text{or,}}\,\,\,{K_2} = 2.25\,K. \cr} $$
Therefore, increase in kinetic energy is $$2.25$$
$$K - K = 1.25\,K\,{\text{or}}\,125\% .$$

For the object of mass $$2.0\,kg.$$
$$\frac{{\Delta k}}{k} = \frac{{k - \frac{k}{4}}}{k} = \frac{3}{4}$$
Kinetic energy transferred
$$\eqalign{ & \frac{{\Delta k}}{k} = \frac{{4{m_1}{m_2}}}{{{{\left( {{m_1} + {m_2}} \right)}^2}}} \cr & {\text{Here,}}\,{m_1} = 2.0\,kg,\;{m_2} = M \cr & \therefore \frac{3}{4} = \frac{{4 \times 2M}}{{{{\left( {2 + M} \right)}^2}}} \Rightarrow M = \frac{2}{3}\,kg\,\,{\text{or}}\,\,6\,kg \cr} $$

Apply conservation of energy. Initially body posseses only kinetic energy and after attaining a height, the kinetic energy is zero.
Therefore, loss of energy $$ = KE - PE$$
$$\eqalign{ & = \frac{1}{2}m{v^2} - mgh \cr & = \frac{1}{2} \times 1 \times 400 - 1 \times 18 \times 10 \cr & = 200 - 180 = 20\,J \cr} $$

$$\eqalign{ & F = v\left( {\frac{{dm}}{{dt}}} \right) = v\frac{d}{{dt}}\left( {\rho \times {\text{Volume}}} \right) \cr & = v\rho \frac{d}{{dt}}\left( {{\text{Volume}}} \right) \cr & = v\rho \times \left( {Av} \right) = A\rho {v^2} \cr & {\text{Power}} = {\text{Force}} \times {\text{Velocity}} \cr & = A\rho {v^2} \times v = A\rho {v^3} \cr & \Rightarrow P \propto {v^3} \cr} $$

As the machine delivers a constant power
$$\eqalign{ & {\text{So }}F \cdot v = {\text{constant}} = k\,\left( {watts} \right) \cr & \Rightarrow m\frac{{dv}}{{dt}} \cdot v = k \cr & \Rightarrow \int v dv = \frac{k}{m}\int d t \cr & \Rightarrow \frac{{{v^2}}}{2} = \frac{k}{m}t \Rightarrow v = \sqrt {\frac{{2k}}{m}t} \cr} $$
Now, force on the particle is given by
$$\eqalign{ & F = m\frac{{dv}}{{dt}} = m\frac{d}{{dt}}{\left( {\frac{{2kt}}{m}} \right)^{\frac{1}{2}}} \cr & = \sqrt {2km} \cdot \left( {\frac{1}{2}{t^{ - \frac{1}{2}}}} \right) = \sqrt {\frac{{mk}}{2}} \cdot {t^{ - \frac{1}{2}}} \cr} $$

The elastic potential energy
$$\eqalign{ & = \frac{1}{2} \times {\text{Force}} \times {\text{Extension}} \cr & = \frac{1}{2} \times 200 \times 0.001 \cr & = 0.1\,J \cr} $$

Speed just before collision $$ = \sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} $$
Speed just after collision $$ = e\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} $$
From conservation of energy,
$$\eqalign{ & \frac{1}{2}m{e^2}\left[ {2g\ell \left( {1 - \cos {\theta _0}} \right)} \right] = mg\ell \left( {1 - \cos \theta } \right) \cr & \therefore \theta = {\cos ^{ - 1}}\left\{ {1 - {e^2}\left( {1 - \cos {\theta _0}} \right)} \right\} \cr} $$

Given, mass of particle $$m = 0.01\,kg.$$
Radius of circle along which particle is moving, $$r = 6.4\,cm.$$
$$\because $$ Kinetic energy of particle, $$KE = 8 \times {10^{ - 4}}J$$
$$\eqalign{ & \Rightarrow \frac{1}{2}m{v^2} = 8 \times {10^{ - 4}}J \cr & \Rightarrow {v^2} = \frac{{16 \times {{10}^{ - 4}}}}{{0.01}} \cr & = 16 \times {10^{ - 2}}\,......\left( {\text{i}} \right) \cr} $$
As it is given that $$KE$$  of particle is equal to $$8 \times {10^{ - 4}}J$$   by the end of second revolution after the beginning of motion of particle. It means, its initial velocity $$\left( u \right)$$ is $$0\,m/s$$  at this moment.
$$\because $$ By Newton's 3rd equation of motion,
$${v^2} = {u^2} + 2{a_t}s$$
$$\eqalign{ & \Rightarrow {v^2} = 2{a_t}S\,\,{\text{or}}\,\,{v^2} = 2{a_t}\left( {4\pi r} \right)\,\,\left( {\because {\text{particle}}\,2\,{\text{revolutions}}} \right) \cr & \Rightarrow {a_t} = \frac{{{v^2}}}{{8\pi r}} = \frac{{16 \times {{10}^{ - 2}}}}{{8 \times 3.14 \times 6.4 \times {{10}^{ - 2}}}}\,\,\left( {\because {\text{from}}\,{\text{Eq}}{\text{.}}\,\left( {\text{i}} \right),{v^2} = 16 \times {{10}^{ - 2}}} \right) \cr & \therefore {a_t} = 0.1\,m/{s^2} \cr} $$