M3 Elastic Springs and Strings Question

Not quite sure how to even start this question - can anyone show me what to do?

A jack-in-the-box is made using a spring of natural length 0.2m and modulus 100N and a 'jack' of mass 0.5kg. When the lid is closed, the spring is compressed to a length of 0.1m. Assuming the spring to be vertical throughout, calculare the maximum distance that the 'jack' will rise when the lid is suddenly raised.

Thank you!

Well you're obviously going to be working with energy (that's what elastics is about). You need to work out the height the jack will reach. Anything come to mind when I present that information that way?

umm im guessing its something to do with EPE=GPE and i know we're trying to find h (as GPE=mgh) but i still cant get the answer

umm im guessing its something to do with EPE=GPE and i know we're trying to find h (as GPE=mgh) but i still cant get the answer

When the spring is compressed to 0.1m, what's the elastic potential energy of it? find it first.

Now when the jack rises, the elastic potential energy is converted into potential energy. The kinetic energy won't be used here, because there is no change in kinetic energy it was zero initially and also at the end.

Hence the equation will be:
loss in epe = gain in pe
EPE is simple to find using the formula $\dfrac{\lambda x^2}{2l}$
PE = mgh = 0.5 x 9.8 x h = 4.9h

Hope it helps .

Basically what raheem said Would +rep you raheem but it won't let me at the moment.

When the spring is compressed to 0.1m, what's the elastic potential energy of it? find it first.

Now when the jack rises, the elastic potential energy is converted into potential energy. The kinetic energy won't be used here, because there is no change in kinetic energy it was zero initially and also at the end.

Hence the equation will be:
loss in epe = gain in pe
EPE is simple to find using the formula $\dfrac{\lambda x^2}{2l}$
PE = mgh = 0.5 x 9.8 x h = 4.9h

Hope it helps .