A body of mass m = 1 kg is dropped from a height h = 40 cm on a horizontal platform fixed to one end of an elastic spring, the other being fixed to a base as shown in the figure. As a result, the spring is compressed by an amount x = 10 cm. What is the force constant of the spring? Take g=10 ms−2.
1000 Nm−1
Since the platform is depressed by an amount x, the total change in potential energy of the body is mg (h + x). This work is stored in the spring in the form of elastic potential energy.
∴12kx2=mg(h+x)⇒k=2mg(h+x)x2
Given, h = 0.4 m, x = 0.1 m, m = 1 kg and g=10 ms−2. Substituting these values, we get k=1000 Nm−1. Hence, the correct choice is (c).