Question

A gymnast of mass 52.0 kg is jumping on a trampoline. She jumps so that her feet reach a maximum height of 2.81 m above the trampoline and, when she lands, her feet stretch the trampoline 67.0 cm down. How far does the trampoline stretch when she stands on it at rest? Assume that the trampoline is described by Hooke's law when it stretched.

Answer 1

$m= 52 \;kg \\ h = 2.81 \;m \\ Δx = 67 \;cm = 0.67 \;m$

The enegry stoored in the trampline for a stretch Δx:

$E = \frac{1}{2}k Δx^2$

k = the spring constant of the trampline

$mgh=\frac{1}{2}k Δx^2 \\ k = \frac{2mgh}{Δx^2} \\ k = \frac{2 \times 52 \times 9.8 \times 2.81}{0.67^2} = 6379.9 \;N/m$

Stretching in the trampoline when the gymnast is standing at rest:

$k(Δx’) = mg \\ Δx’ = \frac{mg}{k} \\ Δx’ = \frac{52 \times 9.8}{6379.9} = 0.07987 \;m = 7.987 \;cm$

Answer: 7.987 cm