Question

A vertical spring with force constant $k$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring, so that the spring is compressed by a distance of $d$. The net work done in the process is
(A) $mg\left( {h + d} \right) + \dfrac{1}{2}k{d^2}$
(B) $mg\left( {h + d} \right) - \dfrac{1}{2}k{d^2}$
(C) $mg\left( {h - d} \right) - \dfrac{1}{2}k{d^2}$
(D) $mg\left( {h - d} \right) + \dfrac{1}{2}k{d^2}$

Answer 1

Hint:The ball falls on the spring and the spring is compressed and you are asked to find the net work that is done in the process. In order to solve this question, you need to consider the forces that will be acting on the ball at various instants or intervals. To find the work done by the forces, take the dot product of the force and displacement. Finally, add up the work by all the forces to find the net work done in the process.

Complete step by step answer:
First, let us analyse the forces acting on the ball. The ball is falling under the action of the gravitational force acting on it by the earth. So, the first force will be forced due to gravity and it will act vertically downward. ${F_g} = mg \downarrow $. This force will act from the start of the motion till its end.The next force that will be acting on the ball will be the force due to spring. As soon as the ball touches the spring, it will apply force in the opposite direction, that is, in the upward direction. ${F_s} = kx \uparrow $, where $x$ is the compression in the spring.Now, let us find the work done by each force.

(1) Due to gravity
The displacement of the ball throughout the process is in the downward direction and also the gravitational force is acting in the downward direction. So, the angle between them is zero.
$d{W_g} = mgdy$
where \[dy\] is the infinitesimal displacement in the vertically downward direction.
$
d{W_g} = {F_g}dy \\
\Rightarrow d{W_g} = mgdy \\
\Rightarrow\int {d{W_g}} = \int\limits_0^{h + d} {mgdy} \\
\Rightarrow{W_g} = mg\left. y \right|_0^{h + d} \\
\Rightarrow{W_g} = mg\left( {h + d} \right) \\
 $

(2) Due to spring
The force on the ball is acting vertically upward and the displacement is happening in a vertically downward direction, so the angle between them is $180^\circ $.
$d{W_s} = - {F_s}dx$, where $dx$ is the infinitesimal compression in the spring in a small interval of time.
$
d{W_s} = - {F_s}dx \\
\Rightarrow d{W_s} = - kxdx \\
\Rightarrow\int {d{W_s} = - \int\limits_0^d {kxdx} } \\
\Rightarrow{W_s} = - k\left. {\dfrac{{{x^2}}}{2}} \right|_0^d \\
\therefore{W_s} = - \dfrac{{k{d^2}}}{2} \\ $
Now, the net work done will be equal to ${W_g} + {W_s}$ which is $mg\left( {h + d} \right) + \left( { - \dfrac{1}{2}k{d^2}} \right) = mg\left( {h + d} \right) - \dfrac{1}{2}k{d^2}$
Therefore, the net work done in the process is $mg\left( {h + d} \right) - \dfrac{1}{2}k{d^2}$

Hence, option B is correct.

Note:In this question, we found out the work done by forces. In general, the work by a force is given as $dW = \overrightarrow F .d\overrightarrow r = Fdr\cos \theta $, where $\theta $ is the angle between force and displacement. In our case, the angle was $0^\circ $ and $180^\circ $, that is why $d{W_g}$ was equal to ${F_g}dy$ and $d{W_s}$ was equal to $ - {F_s}dx$, see the signs. Keep in mind the values of the gravitational force as well as the spring force.