URGENT! HELP!? 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? I have no idea on how to do these...

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1) Resolve for the equilibrium T - mg = 0 T=mg since you know T = ke where e = extension and k =12000 substitute to find "e" 2) The work done by the trampoline in moving 8 cm = Energy stored in the spring F x = 0.5 k x^2 F = 0.5 k x you know k and x...plug in and find AVG force

Can we not use mgx=kx^2/2 to solve the first question?

That looks like the PE = KE equation. That can be used when PE = KE, even though can be true anywhere if you understand that PE is relative to whichever reference you choose. I can explain that if you want. Just ask! The other time it's used is when you say something like \(\rm PE_{before}=KE_{after}\) Which is used when you go from having all energy being only potential energy to a situation where all the potential energy is gone, and it's now all kinetic energy. Side note: this means that we end at the reference of potential energy, where the potential energy equals 0. You can ask about that if you want, too. @thushananth01 used \(T\) to represent the spring force, am I correct? This spring force must equal the wieght of the dude, if the forces cancel out so that the dude can stay still, there. And staying still is what you desire, because the situation is explained to be a dude standing there. Not a dude sinking, or launching into the air. So the methodology is: Look, this guy is standing still. This means no net force, so all forces cancel. So all forces add up to zero (make sure you consider direction). Up force: Spring force Down force: Weight So, if "up" is positive for you, \(\rm Spring\ force+ (-Weight) = 0\) \(\rm Spring\ force-Weight = 0\) \(T-mg=0\) And then you can do algebra from there, as @thushananth01 did.

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