Calculating Spring Constant for an 8kg Stone Resting on a Compressed Spring

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To calculate the spring constant for an 8kg stone resting on a compressed spring, the relationship between gravitational potential energy (Wg) and elastic potential energy (Ws) is crucial, where Wg = -Ws. The spring's compression of 10cm can be used in the formula Ws = -0.5kx^2 to derive the spring constant k. When the stone is pushed down an additional 30cm, the elastic potential energy just before release is calculated as U = 0.5kd^2, yielding a value of 6.27 x 10^-3 J. To find the change in gravitational potential energy as the stone moves to its maximum height, the formula delta(U) = mg[delta(y)] is applied, but the maximum height needs to be determined first.
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An 8kg stone rests on a spring. the spring is compressed 10cm by the stone.

What is the spriong constant?

im not really sure how to start...Ws = -0.5kx^2

im guessing that there must be a relationship between Wg and Ws obviously
-Wg = Ws... but I am not sure how to solve for Ws

Can anyone help?
 
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F = mg = -kx
 
b)the stone is pushed down an extra 30 cm. What is the elastic potential energy of the compressed spring just before the release?

U = 0.5kd^2 d=distance compressed

...U = 6.27 x 10^-3 J

ok...the next part asks for...

what is the change in the gravitational potential energy of the stone-earth system when the stone moves from the release point to its max height

so delta(U) = mg[delta(y)]

how do i get the max height?
 

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