Solve Trampoline Question: Jump from 10m & 20m Height

[vpv]

Homework Statement

A 60 kg person jumps from a platform onto a trampoline 10 m below, stretching it 1.0 m from its initial position. Assumin that the trampoline behaves like a simple elastic spring, how much will it stretch if the same person jumps from a height of 20 m?

Homework Equations

Law of Conservation of Energy
Ek = 0.5mv^2
Eg = mgh
Ee = 0.5kx^2
Fx = kx

The Attempt at a Solution

Really didn't understand the question at all. I figured since the guy is jumping of the platform, velocity at that instance would be 0. If the guy landed on the trampoline, his velocity would be 0 (but i don't know about this). The equillibrium of spring is 0 when the guy jumps of platform so there is no elastic potential. I am guessing there is some gravitational potential as the platform is above the trampoline. But I am consistently getting the wrong answer. The answer is 1.4 m. How would you get that?

Thanks, I appreciate the help.

 

[Hootenanny]

Can you start by writing out an equation representing the principle of conservation of energy?

 

[vpv]

Ek1 + Eg1 + Ee1 = Ek2 +Eg2 +Ee2
0.5mv1^2 + mgh1 + 0.5kx1^2 +FaD = 0.5mv2^2 + mgh2 + 0.5kx2^2 +FfD

 

[Hootenanny]

Ek1 + Eg1 + Ee1 = Ek2 +Eg2 +Ee2
0.5mv1^2 + mgh1 + 0.5kx1^2 +FaD = 0.5mv2^2 + mgh2 + 0.5kx2^2 +FfD

I think you're over-complicating things a little here. If we ignore air resistance, then there are no dissipative forces present and therefore the total energy of the system must remain constant.

So, what type of energy does the system posses before the person jumps and what type of energy does the system posses after the person has landed on the trampoline?

 

[vpv]

I think you're over-complicating things a little here. If we ignore air resistance, then there are no dissipative forces present and therefore the total energy of the system must remain constant.

So, what type of energy does the system posses before the person jumps and what type of energy does the system posses after the person has landed on the trampoline?

Um before the person jumps, he possesses gravitational potential, no? and after he lands, the spring possesses elastic potential.
I think.

 

[Hootenanny]

Um before the person jumps, he possesses gravitational potential, no? and after he lands, the spring possesses elastic potential.
I think.

Correct . So, can you put what you just said in equation form?

 

[vpv]

mgh1 = 0.5kx2^2

but how do i find K?
I tried Fg/x which is 588 N/m. I don't know if that is correct tho Caz using that K value, I get a final answer of 6.3 m which is not even close to the answer at the back. Maybe the answer at the back is wrong?

 

[Hootenanny]

mgh1 = 0.5kx2^2?

That would be correct. Can you now apply this equation to the two cases:

(1) Where the person jumps from 10m
(2) Where the person jumps from 20m

 

[vpv]

That would be correct. Can you now apply this equation to the two cases:

(1) Where the person jumps from 10m
(2) Where the person jumps from 20m

Two words.

You're a Genius.
Wait that's three.

But I want to know why I can't find the k value the way I did. I used Fg/x to find K. Can you tell me what I did wrong?

 

[Hootenanny]

Two words.

You're a Genius.
Wait that's three.

But I want to know why I can't find the k value the way I did. I used Fg/x to find K. Can you tell me what I did wrong?

Note that the force exerted by the spring not only opposes the weight of the trampolinist, but must also account for the acceleration of the trampolinist.

 

[vpv]

AcceleratioN?? How does he accelerate? Do you mean that he decelerates as he lands on the trampoline and his V = 0?

 

[Hootenanny]

AcceleratioN?? How does he accelerate? Do you mean that he decelerates as he lands on the trampoline and his V = 0?

Yes, acceleration is simply a change in velocity - when the speed is increasing or decreasing is irrelevant. Generally in kinematics, one simply talks about acceleration rather than deceleration as things can get a little confusing.