urgent help please?

Question

a vertically supported spring has a force constant of 660n/m. a stone of mass 6kg is placed on top of the spring, pushed down until it is 0.25m from the equilibrum position and the released. how HIGH above the release point does the stone rise ?

Answer 1

The potential energy gained by spring equals 0.5*k*x^2

Here the spring is compressed by 0.25 m

The PE is therefore = 0.2*660*0.25^2 = 20.625 J.

As the spring is released, the potential energy is converted to kinetic energy for the stone.

Once the stone is released, it's kinetic energy is converted to potential. At the maximum height, it possesses only potential energy.

PE = mgh = 20.625

h = 20.625/(9.8*6) = 0.35 m