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 Fig. 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)`.

Options

A. `600 Nm^(-1)`
B. `800 Nm^(-1)`
C. `1000 Nm^(-1)`
D. `1200 Nm^(-1)`

Answer

C

Explanation

Since the platform is depressed by an amount x, the total work done on the spring is mg `(h + x)`. This work is stored in the spring in the form of potential energy `1/2 kx^2`. Equating the two, we have ,
`1/2 kx^2 = mg (h + x) " or " k = (2 mg(h + x))/(x^2) = 1000 Nm^(-1)`