One more question A trampoline and potential energy

[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

 

[Doc Al]

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

Good.

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

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

 

[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

 

[Doc Al]

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...

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.

 

[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

 

[Doc Al]

That looks good.