A 8.00 kg stone is resting on a spring. The spring is compressed 9.0 cm by the stone.?

Question

a) The stone is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release?

b)What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height?

c)What is that maximum height, measured from the release point?

k = 8.7

I need help!

Answer 1

Calculate the k value for the spring. The stone exerts a force of 8*9.8 = 78.4 N and a compression of 9 cm is
78.4/.09 = 871 N/m. (Keep it in the MKS system ☺)

When the spring is pushed down another 30 cm its total deflection is 39 cm and it's total potential energy is
kx²/2 = 871(.39)²/2 = 132.5 J

The stone acquires 132.5 J of gravitational potential energy when released.

Since gravitational potential is mgh, then
8*9.8*h = 132.5 or h = 1.69meters


Doug