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

A ball of mass m is dropped from a heighth on a platform fixed at the

1.	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
Answer» `(MG)/((h + x))` `(mg)/((h + 2x))` `(2mg(h + x))/(x^2)` `(mg)/((2h + x)` SOLUTION :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)` .