100kg person on a pogo stick is landing from 2m (bottom of stick to ground), how much does spring compress?

Question

k=19620 nm
g= 9.81

Answer 1

Before the landing the 100 kg mass compresses the spring an amount x0:

k * x0 = weight = 100 * 9.81

x0 = 981 / k = 981 / 19620
x0 = 0.050 m

This x0 compression stores an initial potential energy U0 in the spring:

U0 = 0.5 * k * x0 ^ 2
U0 = 24.525 J

The Kinetic energy at “the end of the falling” equals the potential energy Ud of the height difference 2 m,

Ud = mgh = 100 * 9.81 * 2
Ud = 1962 J

Max spring compression, x, occurs when the falling kinetic energy Ud and the initial potential energy U0 all stored into the spring:

U0 + Ud = 0.5 * k * x ^ 2
x ^ 2 = (1962 + 24.525) / (0.5 * 19620)
x = 0.450 m

Thus the additional spring compression due to the landing is:
x – x0 = 0.400 m

Note that this 0.400 m additional spring compression is a first order approximation.

The landing height difference used to calculate the 100 kg's potential energy should include this 0.400 m additional downward displacement and not just the 2 m. When first touching the ground it is the moving kinetic energy of the body as a result of the cumulative interaction of the 100 kg mass and the 9.81 gravity field during the 2 m falling that is to be transferred to the potential energy of the spring. An interesting question is whether it is correct to expect the potential energy delta due to this additional 0.400 m downward displacement calculated simply as mgh be also transferred to additional compression potential energy of the spring?