Question

Springs: what is the height bounced by a spring, mass *m*, initial length `l_0` compressed length `l_1`, spring constant *k*?

A pop-up toy consists of a head and sucker of combined mass m stuck to the top of a light spring of natural length `l_0` and spring constant k . The spring is compressed to length `l_1` when the pop-up is stuck to the ground. To what height above the ground does the bottom of the unstretched spring jump to when it is smoothly released?

I know the force of gravity g here is taken as being g. I also reckon its a combination of `E=1/2kx^2` and `E=mgh` and according to a "tips" section on my task also `E=1/2mv^2` though I'm not sure how that fits in.

Any help would be greatly appreciated.

Thank you

Answer 1

So,when you compress the spring by `l_o - l_1`,elastic potential energy stored in it will be `1/2 k (l_o-l_1)^2`

So,the spring due to its restoring force will try to return to its original position and it's energy will get converted to the kinetic energy of the toy i.e `1/2 m u^2` (`u` is the velocity of the toy lying on the top of the spring or we can say the velocity of the top of the unstretched spring,at the beginning point of its journey)

So, `1/2 m u^2 = 1/2 k(l_o-l_1)^2`

So, `u^2 = k/m (l_o-l_1)^2`

Now,if the top of the spring,moves about height `h`,using `v^2 =u^2 -2gh` we get, `0^2 = u^2 -2gh` (as at the maximum height `v=0`)

So, `h= u^2/(2g) = k/(2mg) (l_o-l_1)^2`

Now,the lowest point of the unstretched spring is at `l_o m` distance from its top point,but the top of the unstretched spring was already `l_o m` above the ground.

So,maximum height reached by the bottom of the unstretched spring is `( k/(2mg)(l_o-l_1)^2) -l_o +l_o=k/(2mg) (l_o-l_1)^2`