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

Velocity is maximum when K.E. is maximum.
For minimum P.E.,
$$\eqalign{ & \frac{{dV}}{{dx}} = 0 \cr & \, \Rightarrow {x^3} - x = 0 \cr & \Rightarrow x = \pm 1 \cr & \Rightarrow {\text{Minimum P}}{\text{.E}}{\text{.}} = \frac{1}{4} - \frac{1}{2} = - \frac{1}{4}\,J \cr & {\text{K}}{\text{.E}}{{\text{.}}_{\left( {{\text{max.}}{\text{.}}} \right)}}{\text{ + P}}{\text{.E}}{{\text{.}}_{\left( {\min.} \right)}} = 2\left( {{\text{Given}}} \right) \cr & \therefore {\text{K}}{\text{.E}}{{\text{.}}_{\left( {{\text{max.}}{\text{.}}} \right)}} = 2 + \frac{1}{4} = \frac{9}{4} = \frac{1}{2}mv_{{\text{max.}}{\text{.}}}^2 \cr & \Rightarrow \frac{1}{2} \times 1 \times v_{{\text{max.}}{\text{.}}}^2 = \frac{9}{4} \cr & \Rightarrow {v_{\max .}} = \frac{3}{{\sqrt 2 }} \cr} $$

As $$m$$ is the mass per unit length, then rate of mass per second $$ = \frac{{mx}}{t} = mv$$
∴ Rate of $$KE = \frac{1}{2}\left( {mv} \right){v^2} = \frac{1}{2}m{v^3}$$

$$\eqalign{ & d{U_{\left( x \right)}} = - Fdx \cr & \therefore {U_x} = - \int_0^x {Fdx} = \frac{{k{x^2}}}{2} - \frac{{a{x^4}}}{4} \cr} $$
$$U = 0$$  at $$x = 0$$  and at $$x = \sqrt {\frac{{2k}}{a}} ; \Rightarrow $$   we have potential energy zero twice (out of which one is at origin). Also, when we put $$x = 0$$  in the given function,
we get $$F = 0.\,{\text{But}}\,F = - \frac{{dU}}{{dx}}$$
$$ \Rightarrow {\text{At}}\,x = 0;\frac{{dU}}{{dx}} = 0$$     i.e. the slope of the graph should be zero. These characteristics are represented by (D).

Work done against gravity $$ = mg\sin \theta \times d$$
$$ = 2 \times 10 \times 10 = 200\,J\left( {d\sin \theta = 10} \right)$$
Actual work done $$= 300\,J$$
Work done against friction $$ = 300 - 200 = 100\,J$$

Work done in stretching the rubber-band by a distance $$dx$$   is
$$dW = F\,dx = \left( {ax + b{x^2}} \right)dx$$
Integrating both sides,
$$W = \int\limits_0^L {axdx} + \int\limits_0^L {b{x^2}dx} = \frac{{a{L^2}}}{2} + \frac{{b{L^3}}}{3}$$

Kinetic energy of a body, $$K = \frac{{{p^2}}}{{2m}}$$
$$\eqalign{ & {\text{As}}\,{p_1} = {p_2}\,\left( {{\text{Given}}} \right) \cr & \therefore \frac{{{K_1}}}{{{K_2}}} = \frac{{{m_2}}}{{{m_1}}} = \frac{5}{4} \cr} $$

In elastic collision, kinetic energy of the system remains unchanged and momentum is also conserved.
It is given that mass of balls are same and collision is perfectly elastic $$\left( {e = 1} \right)$$   so their velocities will be interchanged.
$$\eqalign{ & {\text{Thus,}}\,{{v'}_A} = {v_B} = - 0.3\,m/s, \cr & {{v'}_B} = {v_A} = 0.5\,m/s \cr} $$

At equilibrium:
$$\eqalign{ & \frac{{dU\left( x \right)}}{{dx}} = 0 \cr & \Rightarrow \frac{{ - 12a}}{{{x^{11}}}} = \frac{{ - 6b}}{{{x^5}}} \cr & \Rightarrow x = {\left( {\frac{{2a}}{b}} \right)^{\frac{1}{6}}} \cr & \therefore {U_{{\text{at equilibrium}}}} = \frac{a}{{{{\left( {\frac{{2a}}{b}} \right)}^2}}} - \frac{b}{{\left( {\frac{{2a}}{b}} \right)}} = - \frac{{{b^2}}}{{4a}} \cr & {\text{and }}{U_{\left( {x\, = \,\infty } \right)}} = 0 \cr & \therefore D = 0 - \left( { - \frac{{{b^2}}}{{4a}}} \right) = \frac{{{b^2}}}{{4a}} \cr} $$

Work done by friction $$ = \int {\overrightarrow F \cdot \overrightarrow {ds} = } \int\limits_0^x {\mu mg\cos \theta } \frac{{dx}}{{\cos \theta }} = \mu mgx = 20\,J$$

From given graph :
$$\eqalign{ & \vec F = \left( {\frac{3}{4}x + 10} \right)\hat i + \left( {20 - \frac{4}{3}y} \right)\hat j + \left( {\frac{4}{3}z - 16} \right)\hat k \cr & W = \int {\vec F \cdot \overrightarrow {ds} } \cr & = \int\limits_{\left( {0,5,6} \right)}^{\left( {2,10,0} \right)} {\left( {\frac{3}{4}x + 10} \right)\hat i + \left( {20 - \frac{4}{3}y} \right)\hat j + \left( {\frac{4}{3}z - 16} \right)\hat k} \cr & \left[ {dx\,\hat i + dy\,\hat j + dz\,\hat k} \right] = \frac{{287}}{2}J \cr} $$
Work done can also be found by finding area under these curves.