# One more question A trampoline and potential energy

1. May 28, 2007

### MJC8719

A 75 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 4.0 m/s. the tampoline is three meters below the platform.

(b) If the trampoline behaves like a spring of spring constant 5.2 104 N/m, how far does he depress it?

when he depressed the trampoline he also moved further downward. So the total change in height is H + x where H = 3.0 meters and x is the distance the trampoline is depressed, not just 3 meters

You can equate energy at the top, when he is motionless, with energy at the bottom, when he is also motionless.

So equating energy...

m g (H + x) = (1/2) k x2

You'll have a quadratic in x. Go ahead and plug in numbers:

75 * 9.8 * ( 3 + x ) = (1/2) * 52000 x2

2205 + 735 x = 26000 x2 ugly. you can div everything by 1000

26 x2 - 0.735 x - 2.205 = 0 and use quadratic eqn

x = ( 0.735 +/- (0.7352 - 4* 26* (-2.205) )1/2 ) / 2*26

x = ( 0.735 +/- 15.16 ) / 52 = -0.277, 0.306

The negative is not valid, so the answer must be 0.306 meters

2. May 28, 2007

### Staff: Mentor

Good.

But he's not motionless when he jumps from the platform; he's moving at 4.0 m/s.

3. May 28, 2007

### MJC8719

Ok...I understand that I missed that but I do not know how I can account for it.....Is it as simple as multiplying the mg(h+x) by 4 or is it more complicated....

Im guesssing I ignored that part since I am not to sure what to do with it lol

4. May 28, 2007

### Staff: Mentor

mg(h+x) is potential energy; why would you multiply an energy times a velocity???

Instead, realize that the initial energy is not just gravitational PE, but also KE.

5. May 28, 2007

### MJC8719

Whoops..

So the original equation should read:
mg(h+x) + .5mv^2 (accounting for the kinetic energy) = .5 kx^2

Then just solve for x

6. May 28, 2007

### Staff: Mentor

That looks good.