# Vertical spring (trampoline) compresion.

**1. The problem statement, all variables and given/known data**

A student jumping on a trampoline reaches a maximum height of h = 0.96 m. The student has a mass of m = 58 kg.

What is the student’s speed immediately before she reaches the trampoline after the jump in m/s?

**answer:**4.338

If, when she lands on the trampoline, she stretches the trampoline down d = 0.75 m, what is the spring constant k in N/m of the trampoline?

h=0.96m

m=58 kg

d= 0.75

v=4.338

**2. Relevant equations**

KE=(1/2)mv

^{2}

F=(1/2)kx

SpringPE=(1/2)kx

^{2 }

ΔPe=mgh

W=fd cos(Θ)

KE=-PE

**3. The attempt at a solution**

I have had a gruesome time trying to understand spring related questions.

First I took F=-kx and solved for k, leading to -k=F/x leading to k=mg/x. This is wrong however, because the force of the trampoline is greater than the force exerted by gravity on the person. so in the equation we have 2 unknowns, k and f.

So I took formula KE=-PE

KE=(1/2)mv^{2}

SpringPE=(1/2)kx^{2}

leading to (1/2)mv^{2}=-(1/2)kx^{2}

thus -mv^{2}/x^{2}=k

This didn’t give the correct answer and I’m not sure why. Is KE also supposed to inclued the KE of the falling person?

http://ift.tt/1gdQvq9

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