# Homework Help: Spring-like trampoline

1. Dec 4, 2013

### lollikey

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

A spring-like trampoline dips down 0.08 m when a particular person stands on it. If this person jumps up to a height of 0.26 m above the top of the trampoline, how far with the trampoline compress when the person lands?

2. Relevant equations

F=kx

U= .5 * k * x^2

PE= mgh

3. The attempt at a solution

PE = U

m * 9.8 * .18 = .5 * k * .18^2

2. Dec 4, 2013

### rock.freak667

Well initially when the person is standing, weight = spring force or mg = (0.08)k.

So mg/k = 0.8

Now in mgh = 1/2 kx ^2, you can divide by k.

3. Dec 4, 2013

### lollikey

.08k = .5 * k * x^2

.08= .5 x^2

.16= x^2

x= .40

but it marked it wrong

4. Dec 4, 2013

### rock.freak667

Because you are equating spring force to energy

Your equation is PE = U or mgh= ½kx2

5. Dec 4, 2013

### lollikey

.08k *.26 = .5 * k * x^2
.0208 = .5 * x^2
.0146 = x^2
x = .2039

am I using the wrong height?

.08k * h = .5 * k * .26^2
.08 * h = .0338
h = .4225

none of these are right

Last edited: Dec 4, 2013
6. Dec 5, 2013

### rock.freak667

Right initially the spring is compressed 0.08 m so that mg = 0.08k

Now as the person falls, they will travel a distance 'h' to the point where they just hit the trampoline and they will continue to now compress the spring a distance 'x'. The total of these are then converted into the elastic potential energy of the spring.

So now you have mgh + mgx = ½kx2.

You will have a quadratic to solve in x.