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

$$\eqalign{ & {\text{Here,}}\,m = 4,g = 4 \times {10^{ - 3}}\,kg \cr & x = 4{t^2} + t \cr & \therefore \frac{{dx}}{{dt}} = 8t + 1\,\,\frac{{{d^2}x}}{{d{t^2}}} = 8 \cr & {\text{Work}}\,{\text{done,}}\,\,W = \int f dx = \int m \frac{{{d^2}x}}{{d{t^2}}}\left( {\frac{{dx}}{{dt}}} \right)dt \cr & = \int\limits_0^2 {\left( {4 \times {{10}^{ - 3}}} \right)} \left( 8 \right)\left( {8t + 1} \right)dt \cr & = 32 \times {10^{ - 3}}\int\limits_0^2 {\left( {8t + 1} \right)} dt = 32 \times {10^{ - 3}}\left[ {\frac{{8{t^2}}}{2} + t} \right]_0^2 \cr & = 32 \times {10^{ - 3}}\left[ {4{{\left( 2 \right)}^2} + 2 - 0} \right] = 576\,mJ \cr} $$

Let acceleration of body be a
$$\eqalign{ & \therefore {v_1} = 0 + a{t_1} \Rightarrow a = \frac{{{v_1}}}{{{t_1}}} \cr & \therefore v = at \Rightarrow v = \frac{{{v_1}t}}{{{t_1}}} \cr & {P_{{\text{inst}}}} = \vec F.\vec v = \left( {m\vec a} \right).\vec v \cr & = \left( {\frac{{m{v_1}}}{{{t_1}}}} \right)\left( {\frac{{{v_1}t}}{{{t_1}}}} \right) = m{\left( {\frac{{{v_1}}}{{{t_1}}}} \right)^2}t \cr} $$

The centripetal acceleration
$$\eqalign{ & {a_c} = {k^2}\,r\,{t^2}\,\,\, \Rightarrow \frac{{{v^2}}}{r} = {k^2}\,r\,{t^2} \cr & \Rightarrow \frac{1}{2}m{v^2} = \frac{m}{2}{k^2}\,{r^2}\,{t^2}.....(i) \cr & \Rightarrow K.E. = \frac{m}{2}{k^2}\,{r^2}\,{t^2} \cr & \Rightarrow \frac{d}{{dt}}K.E. = m\,{k^2}\,{r^2}\,t \cr & \Rightarrow {\text{Power}} = m\,{k^2}\,{r^2}\,t \cr} $$

Let $${p_1}$$ be the initial momentum and $${p_2}$$ be the inversed momentum
So, $${p_2} = \frac{{150}}{{100}}{p_1}$$
i.e. \[m{v_2} = \frac{{15}}{{10}}m{v_1}\,\,\left( {\begin{array}{*{20}{c}} {{p_1} = m{v_1}}\\ {{p_2} = m{v_2}} \end{array}} \right)\]
or $${v_2} = \frac{{15}}{{10}}{v_1}$$
Now, $$\frac{{{E_2}}}{{{E_1}}} = \frac{{\frac{1}{2}mv_2^2}}{{\frac{1}{2}mv_1^2}}$$
$$ = {\left( {\frac{{{v_2}}}{{{v_1}}}} \right)^2} = {\left( {\frac{{15}}{{10}}} \right)^2} = \frac{{225}}{{100}}$$
Clearly, $${E_2} > {E_1}$$
So, percentage increase in $$KE$$
$$\eqalign{ & = \frac{{\left( {{E_2} - {E_1}} \right)}}{{{E_1}}} \times 100 \cr & = \left( {\frac{{225}}{{100}} - 1} \right) \times 100 \cr & = 125\% \cr} $$

The expression of work done by the variable force $$F$$ on the particle is given by
$$W = \int {\overrightarrow F \cdot \overrightarrow {d\ell } } $$
In going from $$\left( {0,0} \right)$$  to $$\left( {a,0} \right),$$  the coordinate of $$x$$ varies from 0 to $$'a',$$ while that of $$y$$ remains zero. Hence, the work done along this path is :
$${W_1} = \int_0^a {\left( { - kx\hat j} \right) \cdot dx\hat i = 0\,\,\left[ {\because \hat j \cdot \hat i = 0} \right]} $$
In going from $$\left( {a,0} \right)$$  to $$\left( {a,a} \right)$$  the coordinate of $$x$$ remains constant $$\left( { = a} \right)$$  while that of $$y$$ changes from $$0$$ to $$'a'.$$ Hence, the work done along this path is
$${W_2} = \int_0^a {\left[ {\left( { - K} \right.\left( {y\hat i + a\hat j} \right) \cdot dy\hat j} \right]} = ka\int_0^a {dy = - K{a^2}} $$

Hence, $$W = {W_1} + {W_2} = - K{a^2}$$

Initial kinetic energy of the system
$$K.{E_i} = \frac{1}{2}m{u^2} + \frac{1}{2}M{\left( 0 \right)^2} = \frac{1}{2} \times 0.5 \times 2 \times 2 + 0 = 1\,J$$
For collision, applying conservation of linear momentum $$m \times u = \left( {m + M} \right) \times v$$
$$\eqalign{ & \therefore 0.5 \times 2 = \left( {0.5 + 1} \right) \times v \cr & \Rightarrow v = \frac{2}{3}\,m/s \cr} $$
Final kinetic energy of the system is
$$K.{E_f} = \frac{1}{2}\left( {m + M} \right){v^2} = \frac{1}{2}\left( {0.5 + 1} \right) \times \frac{2}{3} \times \frac{2}{3} = \frac{1}{3}\,J$$
$$\therefore $$ Energy loss during collision $$ = \left( {1 - \frac{1}{3}} \right)J = 0.67\,J$$

Given force $$\vec F = 2t\hat i + 3{t^2}\hat j$$
According to Newton's second law of motion,
$$\eqalign{ & m\frac{{d\vec v}}{{dt}} = 2t\hat i + 3{t^2}\hat j\,\,\left( {m = 1\,kg} \right) \cr & \Rightarrow \int\limits_0^{\vec v} {d\vec v} = \int\limits_0^t {\left( {2t\hat i + 3{t^2}\hat j} \right)dt} \Rightarrow \vec v = {t^2}\hat i + {t^3}\hat j \cr & {\text{Power}}\,P = \vec F \cdot \vec v\left( {2t\hat i + 3{t^2}\hat j} \right) \cdot \left( {{t^2}\hat i + {t^3}\hat j} \right) \cr & = \left( {2{t^3} + 3{t^5}} \right)W \cr} $$

Applying momentum conservation
$$\eqalign{ & {m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2} \cr & 0.1u + m\left( 0 \right) = 0.1\left( 0 \right) + m\left( 3 \right) \cr & 0.1u = 3m \cr & \frac{1}{2}0.1{u^2} = \frac{1}{2}m{\left( 3 \right)^2} \cr} $$
Solving we get, $$u = 3$$
$$\eqalign{ & \frac{1}{2}k{x^2} = \frac{1}{2}K{\left( {\frac{x}{2}} \right)^2} + \frac{1}{2}\left( {0.1} \right){3^2} \cr & \Rightarrow \frac{3}{4}k{x^2} = 0.9 \Rightarrow \frac{3}{2} \times \frac{1}{2}k{x^2} = 0.9 \cr & \therefore \quad \frac{1}{2}K{x^2} = 0.6\,J\,\left( {{\text{total initial energy of the}}\,{\text{spring}}} \right) \cr} $$

$$\eqalign{ & d{U_{\left( x \right)}} = - Fdx \cr & \therefore {U_x} = - \int_0^x {Fdx} = \frac{{k{x^2}}}{2} - \frac{{a{x^4}}}{4} \cr} $$
$$U=0$$  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 $$F=0.$$   But $$F = - \frac{{dU}}{{dx}}$$
$$ \Rightarrow $$ At $$x = 0;\,\,\,\frac{{dU}}{{dx}} = 0$$     i.e., the slope of the graph should be zero. These characteristics are represented by (D).

Let $$u$$ be the speed with which the ball of mass $$m$$ is projected.
Then the kinetic energy $$\left( E \right)$$ at the point of projection is
$$E = \frac{1}{2}m{u^2}\,......\left( {\text{i}} \right)$$
When the ball is at the highest point of its flight, the speed of the ball is $$\frac{u}{{\sqrt 2 }}$$ (Remember that the horizontal component of velocity does not change during a projectile motion).
$$\therefore $$ The kinetic energy at the highest point
$$ = \frac{1}{2}m{\left( {\frac{u}{{\sqrt 2 }}} \right)^2} = \frac{1}{2}\frac{{m{u^2}}}{2} = \frac{E}{2}\,\,\left[ {{\text{From}}\,\left( {\text{i}} \right)} \right]$$