Posted by **Pink ** on Friday, April 18, 2014 at 10:11pm.

1) Determine the extension of the trampoline when a 200 kg person stands on the trampoline. (k=12000)

2) If on the first push, the trampoline went 8.0 cm below its position in part 1, find the average force exerted by his leg muscles?

For the first question i was able to get 0.33m (i think thats the answer, i used mgx=1/2kx^2)

but i have no idea how to do the second question...

- Physics -
**bobpursley**, Friday, April 18, 2014 at 10:48pm
1. wrong equation.

Hookes law: force=mg=kx

2. force*distance=1/2 kx^2

Now really, think on that, it gives you half the force on part 1. But because force is linear, from zero to max, it makes sense that the average force then is half the max force, kx.

Avg force=1/2 kx

- Physics -
**Pink **, Friday, April 18, 2014 at 10:59pm
Okay, so I understand how to do the first question now, but i'm still not clear on how you did the second one...

- Physics -
**bobpursley**, Friday, April 18, 2014 at 11:04pm
work going into trampolene= avg force*distance= avgF*x

that equals the stored enrgy 1/2 kx^2

avgF=1/2 kx

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