Question

A gymnast of mass 51.00 kg is jumping on a trampoline. She jumps so that her feet reach a maximum height of 2.100 m above the trampoline and, when she lands, her feet stretch the trampoline down 61.00 cm. How far does the trampoline stretch when she stands on it at rest? [Hint: Assume the trampoline obeys Hooke's law when it is stretched.]

Answer 1

Answer:

$x = 6.7 cm$

Explanation:

Here we can use energy conservation for two positions of the gymnast

When she is at the lowest position of the trampoline then its potential energy will convert into gravitational potential energy at the top

So we will have

$\frac{1}{2}kx^2 = mg(H + x)$

so we have

$\frac{1}{2}k(0.61^2) = 51 \times 9.81(2.1 + 0.61)$

$k = 7287.5$

now when she stands on it

then by force balance we will have

$mg = kx$

$51 \times 9.81 = 7287.5 \times x$

$x = 6.7 cm$