Question

If a 40 kg gymnast and a 400 kg sumo wrestler each dropped from 1 m above the trampoline, find the final position of each athlete. Assume the trampoline is a simple spring obeying Hooke's law with a k value of 12 000 N/m.

Answer 1

When you drop an object from some height on the spring objects initial potential energy is converted to kinetic energy, this kinetic energy is then used to compress the spring. Once all the energy is used the spring stops compressing and starts oscillating. 
We need to find how much the spring compressed in both cases.
From the above analysis, we can conclude that potential energy in the gravitational field has to be equal to the potential energy of compressed spring.
$mgh=\frac{1}{2}kx^2$
We solve for x (in our case h=1):
$mg=\frac{1}{2}kx^2\\ x^2=\frac{2mg}{k}\\ x=\sqrt{\frac{2mg}{k}}$
Now we just have to plug in the numbers:
$Sumo\ wrestler(m=400kg):x=\sqrt{\frac{2mg}{k}}=0.81m$
$Gymnast(m=40kg):x=\sqrt{\frac{2mg}{k}}=0.26m$