Vertical spring (trampoline) compresion.

• Sane Co.
In summary: 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=-PEKE=(1/2)mv2SpringPE=(1/2)kx2leading to (1/2)mv2=-(1/2)kx2thus -mv2/x
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?
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
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?

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

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
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?

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.

(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?

1. What is a vertical spring (trampoline) compression?

A vertical spring (trampoline) compression is a type of motion that occurs when a person jumps or bounces on a trampoline, causing the spring to compress and then release, propelling the person into the air.

2. How does a vertical spring (trampoline) compression work?

When a person jumps on a trampoline, their weight causes the springs to compress, storing potential energy. As they reach the bottom of their jump, the springs release the stored energy, propelling the person back into the air.

3. What factors affect the compression of a vertical spring (trampoline)?

The factors that affect the compression of a vertical spring (trampoline) include the weight and force of the person jumping, the strength and tension of the springs, and the shape and size of the trampoline.

4. Is vertical spring (trampoline) compression the same as elasticity?

No, vertical spring (trampoline) compression is not the same as elasticity. Elasticity refers to the ability of a material to return to its original shape after being stretched or compressed, while vertical spring compression specifically describes the motion of a trampoline's springs under the influence of an external force.

5. How does vertical spring (trampoline) compression relate to physics?

Vertical spring (trampoline) compression is related to physics because it involves the principles of energy, force, and motion. The compression of the springs and the subsequent release of energy can be explained using concepts such as potential and kinetic energy, and the laws of motion.

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