# Homework Help: Jumping on a Spring-Like Trampoline problem

1. Jul 26, 2014

### KKazaniecki

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

A spring-like trampoline dips down 0.07 m when a particular person stands on it. If this person jumps up to a height of 0.35 m above the top of the uncompressed trampoline, how far will the trampoline compress after the person lands?

2. Relevant equations

here are the relevant equations I could think of :

We could use: the potential energy for the spring : U=1/2kx^2
Gravitational potential ( near earth ) : U=mgh
Kinetic energy : K = 1/2mv^2
Hooke's law: F=-kx

3. The attempt at a solution

btw , before you start : I'm using x for the axis of up and down . Because the above equations have x in them and I forgot to change it to y , as it traditionally should be.

hey , this problem popped up in my mechanics homework this week. So here's my attempt at it. But somehow when I enter the answer I get it as wrong.

well okay . So first thing I guess we're gonna have to do is find the Spring Constant.

which I found to be . mg/xi . I got this by using the fact that when it's stationary. The forces cancel out. So using hooke's law: mg ( force pointing downwards ) is equal to -kxi . where k is the spring constant and x is how much the spring is the displacement of the spring form the relaxed point . Re arranging mg=-kx we get k=mg/x . And here I drop the negative sign because the we're only dealing with magnitudes here. Right so we plug in the numbers and we get k = 60*9.8/0.07 which is 8400. btw we're just giving the person weight of 60kg but could be anything else.

ok so K = 8400.

now we can use other equations to find the answer.

so, When the person jumps and reaches it's maximum height of 0.35 meters in this case. He doesn't move for a moment. So his kinetic energy at that point is 0. Meaning all the energy is potential . And we have the Gravitational potential equation. Which is U= mgh. So then as it falls. it gains kinetic energy etc etc. And since there is no external forces in action. the energy is conseved. And the Potential Energy the spring ends up being equal to the potential gravitational energy. Which we get . Spring = Gravity , 1/2kx2=mgh . and rearranging that we get
x = sqrt((2mgh)/k)

right so , plugging in the numbers we obtain x = sqrt((2*60*9.8*0.35)/8400) = 0.2213594362.

So x=0.2213594362 . and I input that number and I get marked wrong. So I try rounding it to 2 decimal places , since all the givens are to 2 decimal places. But I still get marked wrong. so , please could someone check my process , and my answer and point out where I went wrong. or something. Because it's really doing my head in , the method I'm using seems to right ( well at least for me hehe ). Anyways any help would be highly appreciated :) have a great day people
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 26, 2014

### Nathanael

By your logic, if the initial height (and therefore initial gravitational potential energy) is zero, then the spring will not be compressed at all

Last edited: Jul 26, 2014
3. Jul 26, 2014

### KKazaniecki

what I meant is that . It's one or the other. So the energy potential at the maximum height is the same as the energy potential of the spring when it is compressed. Because Energy is conserved . It goes from Gravitational Potential --> to kinetic ( as it falls ) ---> to spring potential ( when compressed ). Since there is no friction or air resistance there is no change in the Total Energy. Is what I'm saying true or have I missed something. Thanks.

4. Jul 26, 2014

### Nathanael

Right, but you only measured the (change in) Gravitational Potential Energy from the height that it initially was to the height of the unstretched spring.
What about the gravitational potential energy gained while the spring is being compressed?

5. Jul 27, 2014

### KKazaniecki

ok so , mgh + mgx = 1/2kx^2 . In other words Gravitational Potential = Spring Potential. Which as you said is the one where it's above the spring , and when it's compressing the spring. So Energy is conserved .
Rearranging that to find x , We get: x = (sqrt(g m (g m+2 h k))+g m)/k. Where k = gm/0.07. ( hooke's law ). And So x = 0.30216373532 . Rounding that to 2 d.p we get x = 0.30. And I just checked it and it's right. Thanks so much

6. Jan 23, 2015