Work Energy And Power MCQs for NEET — Physics Questions with Answers

Key Idea        The energy lost due to air friction is equal to difference of initial kinetic energy and final potential energy.

      Initially, body posses only kinetic energy and after attaining a height the kinetic energy is zero.

      Therefore, loss of energy=KE-PE

                       $$\begin{gathered} =\frac{1}{2} {\mathrm{mv}}^2-\mathrm{mgh} \\ =\frac{1}{2}\times 1\times 400-1\times 18\times 10 \\ =200-180 \\ =20 \mathrm{J} \end{gathered}$$

Use the law of conservation of energy. Let x be the extension in the spring.

Applying conservation of energy-

                     $mgx-\frac{1}{2}k{x}^2=0-0$

  $\Rightarrow x=\frac{2 mg}{k}$

(d)

 $$\begin{gathered} At extension l_1 , the stored energy =\frac{1}{2}K{l}_1^2 \\ At extension l_2 , the stored energy =\frac{1}{2}K{l}_2^2 \\ Work done in increa\sin g its extension from l_1 to l_2 \\ =\frac{1}{2}K\left(l_2^2-l_1^2\right) \end{gathered}$$

$K_i=\frac{1}{2}m{u}^2$

By conservation of momentum mu = $m{v}_1+nm{v}_2$

$$\begin{gathered} u=v_1+n{v}_2 \left(1\right) \\ {v}_2-v_1=u_1-u_2 \left(2\right) \\ \therefore v_2-v_1=u \\ \left(1\right)+\left(2\right) \\ 2u=(n+1)v_2 \\ {v}_2=\frac{2u}{n+1} \end{gathered}$$

Kinetic energy transferred $$\begin{gathered} K_t=\frac{1}{2}\left(nm\right)v_2^2, K_t=\frac{2nm{u}^2}{{(n+1)}^2}, \\ \frac{K_t}{K_i}=\frac{4n}{{(n+1)}^2} \end{gathered}$$

Between elastic collision of two identical masses, velocities are interchanged.

$\begin{array}{rl}\text{Kinetic energy }\mathrm{E}& =\frac{{\mathrm{p}}^2}{2\mathrm{m}}\\ & =\frac{(\mathrm{Ft}{)}^2}{2\mathrm{m}}=\frac{{\mathrm{F}}^2{\mathrm{t}}^2}{2\mathrm{m}}\end{array} \text{(As }\mathrm{p}=\mathrm{Ft})$

According to the NCERT text, 'Every force encountered in mechanics does not have an associated potential energy. For example, work done by friction over a closed path is not zero and no potential energy can be associated with friction.' Gravitational, spring, and electrostatic forces are all conservative forces and thus have associated potential energy functions.

The NCERT text states: 'If the two forces on the body consist of a conservative force $F_c$ and a non-conservative force $F_{nc}$, the conservation of mechanical energy formula will have to be modified. By the WE theorem $(F_c + F_{nc}) \Delta x = \Delta K$. But $F_c \Delta x = - \Delta V$. Hence, $\Delta(K + V) = F_{nc} \Delta x$, which means $\Delta E = F_{nc} \Delta x$. Over the path this assumes the form $E_f - E_i = W_{nc}$ where $W_{nc}$ is the total work done by the non-conservative forces over the path.' Therefore, the total mechanical energy changes by an amount equal to the work done by the non-conservative force.

The NCERT text states that for a conservative force, 'The work done by the conservative force depends only on the end points.' It also implies that if the work done or kinetic energy did depend on factors like path taken, 'the force would be called non-conservative.' Thus, for a non-conservative force, the work done depends on the path taken.