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NEET PHYSICSEasy

A vertical spring with a force constant kk is fixed on a table. A ball of mass mm at a height hh above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance dd. The net work done in the process is:

A

mg(h+d)+12kd2mg(h+d) + \frac{1}{2}kd^2

B

mg(h+d)12kd2mg(h+d) - \frac{1}{2}kd^2

C

mg(hd)12kd2mg(h-d) - \frac{1}{2}kd^2

D

mg(hd)+12kd2mg(h-d) + \frac{1}{2}kd^2

Step-by-Step Solution

The total work done on the ball during this process is the algebraic sum of the work done by the gravitational force and the work done by the spring force.

  1. Work done by Gravity (WgW_g): The force of gravity (mgmg) acts vertically downwards. The ball moves from an initial height hh above the spring and further descends a distance dd as the spring compresses. The total vertical displacement in the direction of the force is (h+d)(h+d). Thus, the work done by gravity is positive: Wg=mg(h+d)W_g = mg(h+d).
  2. Work done by the Spring Force (WsW_s): As the spring is compressed, it exerts a restoring force Fs=kxF_s = -kx in the upward direction, opposing the downward displacement dd. Since the force and displacement are in opposite directions, the work done by the spring is negative. Following the formula for spring work, Ws=0dkxdx=12kd2W_s = -\int_0^d kx dx = -\frac{1}{2}kd^2.
  3. Net Work Done (WnetW_{net}): Summing the individual work components gives: Wnet=Wg+Ws=mg(h+d)12kd2W_{net} = W_g + W_s = mg(h+d) - \frac{1}{2}kd^2.
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