Vertical spring (trampoline) compresion.

[Sane Co.]

Homework Statement

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

Homework Equations

KE=(1/2)mv²
F=(1/2)kx
SpringPE=(1/2)kx²
ΔPe=mgh
W=fd cos(Θ)
KE=-PE

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)mv²
SpringPE=(1/2)kx²
leading to (1/2)mv²=-(1/2)kx²
thus -mv²/x²=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?

 

[dauto]

seems like you forgot to include gravitational potential energy in your calculation

 

[Simon Bridge]

When the student reaches maximum height, is she in contact with the surface of the trampoline?

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

They can be pretty gruesome.

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.

Yes: mg=kx when the person is just standing on the trampoline - this is the equilibrium condition.
When she jumps, she passes through the equilibrium.

So I took formula KE=-PE
KE=(1/2)mv2
SpringPE=(1/2)kx2
leading to (1/2)mv2=-(1/2)kx2
thus -mv2/x2=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?

Where else would you get the KE from?
Why is your value of k negative? Does that make sense?

Your calculation assumes that all the KE from the person falling a height h goes into the trampoline - and nothing else. What about gravitational PE dropping additional distance d?

 

[Sane Co.]

Thank you guys very much. With your help and suggestions, I have come up with the equation (1/2)mv^2=0.5kx^2+mgh.
I thank you guys very much.

 

[Simon Bridge]

(1/2)mv^2=0.5kx^2+mgh

Um...

You should be careful to use the variables from the problem statement in your formulas.
i.e. (1/2)mv^2 is the kinetic energy gained falling a distance h ... which is mgh
so you wrote: mgh=0.5kx^2+mgh which means that k=0.

I'm sure that is not what you meant?

Note: how much of that initial GPE ends up stored in the spring?