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 $d$ . The net work done in the process is

• A. $mg(h+d)-\frac{1}{2}kd^2$
• B. $mg(h-d)-\frac{1}{2}kd^2$
• C. $mg(h-d)+\frac{1}{2}kd^2$
• D. $mg(h+d)+\frac{1}{2}kd^2$

Answer 1

The Correct Option is A

To calculate the net work done when a ball of mass $m$ falls on a vertical spring and compresses it by distance $d$, we use the principle of conservation of energy. Here is the step-by-step solution:

1. \text{Initial gravitational potential energy of the ball} = mgh
2. \text{Work done by the ball in compressing the spring by } d = \frac{1}{2}kd^2
3. \text{Final gravitational potential energy when the ball compresses the spring by } d = mg(h+d) , because the ball moves downward an additional distance d besides the initial height h.
4. According to the work-energy principle, the net work done by external forces is equal to the change in mechanical energy:
   \text{Net work done} = \text{Initial Potential Energy} - \left(\text{Final Potential Energy} + \text{Energy stored in the spring}\right)
5. Substituting the values, we get:
   \text{Net work done} = mgh - \left(mg(h+d) + \frac{1}{2}kd^2\right)
6. Simplifying,
   \text{Net work done} = mgh - mgh - mgd - \frac{1}{2}kd^2 = -mgd - \frac{1}{2}kd^2
7. Therefore, the correct expression for the net work done is
   mg(h+d) - \frac{1}{2}kd^2

Thus, $mg(h+d)-\frac{1}{2}kd^2$ is the correct answer.