Question

A young boy of mass m = 25 kg sits on a coiled spring that has been compressed to a length 0.4 m shorter than its uncompressed length and then held at this length. Suddenly the spring is released, and the boy flies vertically into the air. He reaches a maximum distance 0.5 m above his initial position. The spring is ideal and massless and we ignore the air friction.

What is the spring constant k of the spring?

What is the speed of the boy when he is 0.4 meters above his starting position?

Answer 1

The spring's elastic potential energy is converted into elastic potential energy of the boy.
Potential energy = elastic energy
mgh = 1/2 kx²
k = (2 * 25 * 9.81 * 0.5) / 0.4²
k = 1533 N/m

We will apply
2as = v² - u², with u = 0
a = F/m
F = kx
a = kx/m

v = √(2 x 0.4 x (0.4 x 1533 / 25))
v = 4.43 m/s