A ball of mass m is dropped from a heighth on a platform fixed at the top of a vertical spring, as shown in figure. The platform is depressed by a distance x. Then the spring constant is 

Options

A. `(mg)/((h + x))`
B. `(mg)/((h + 2x))`
C. `(2mg(h + x))/(x^2)`
D. `(mg)/((2h + x)`

Answer

C

Explanation

Loss in potential energy of the ball = `mg(h + x)`
Gain in elastic potential energy of the spring = `1/2 kx^2`
According to the law of conservation of mechanical energy
`mg(h + x) = 1/2 kx^2 " or " k = (2mg (h + x))/(x^2)` .