Asked by Pink
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 that's the answer, i used mgx=1/2kx^2)
but i have no idea how to do the second question...
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 that's the answer, i used mgx=1/2kx^2)
but i have no idea how to do the second question...
Answers
Answered by
bobpursley
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
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
Answered by
Pink
Okay, so I understand how to do the first question now, but i'm still not clear on how you did the second one...
Answered by
bobpursley
work going into trampolene= avg force*distance= avgF*x
that equals the stored enrgy 1/2 kx^2
avgF=1/2 kx
that equals the stored enrgy 1/2 kx^2
avgF=1/2 kx
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