Solving Physics Problem: Calculating Net Stretch for 55.3kg Person

[ninjagowoowoo]

Q:
A 55.3 kg person jumps from a window to a fire net 21.9 m below, which stretches the net 1.04 m. Assume that the net behaves like a simple spring, and calculate how much it would stretch if the same person were lying in it.

Here's what I did:

using conservation of energy to get the velocity of the person right before it hits the net. (mgh = 0.5mv^2) Once I got the velocity, I found the acceleration caused by the net (deceleration if you will) (v^2=v^2 + 2ay). Then using that acceleration, I found the force of the net on the person (F=ma) Once I found that force, I used it to find the spring constant(k) of the "spring". (F=ky). Finally, using that k, I found the displacement of the "spring" caused by the force of just the person's weight (F=kx where F=55.3kg(9.8)) and got an answer, which was wrong.

Can anyone help me?

 

[Berislav]

You forgot to take into consideration the gravitational potential energy difference while the person was on the trampolin.

So,

\frac{kx^2}{2}=mg(h+x)

 

[ninjagowoowoo]

hmm I don't quite understand the formula you came up with. I understand what you are saying, but perhaps you could elaborate a little more on how you came up with that formula. Thanks.

 

[ninjagowoowoo]

oh by the way, thanks for the help.

 

[Berislav]

Law of conservation of energy. The gravitational potential energy is converted into the energy of the spring.

E=\frac{kx^2}{2}

The person doesn't only fall 21.9 m, he also falls for an additional 1.04 m while he's streching the trampoline.