## anonymous 2 years ago 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. anonymous

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

2. anonymous

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

3. theEric

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.

4. anonymous

Oh okay thank you. I think I understand now. So it is because the person is stationary that we had to use the T-mg=0 method.

5. theEric

Right! The total force must be 0. :)

6. anonymous

Haha, now it it sounds simple. But I still don't understand how the second question was answered by @thushananth01

7. theEric

Okay! Well, now you want to think about potential energy of the spring. Are you comfortable with that? $${\rm PE}=\frac12kx^2$$

8. anonymous

yes

9. theEric

Okay. So now we want to think about work! That is, force times displacement if they're in the same direction. Like, $$W=F\dot\ d\cos\theta$$ or, here, $$W=F\dot\ d\cos(90^\circ)=F\dot \ d$$ Are you okay with work?

10. theEric

I guess I should ask if you're comfortable with that definition of work.

11. anonymous

yup im comfortable with all the basic energy stuff, including gravitational, kinetic, elastic, and work. I just can't seem to know how to apply them for this question.

12. theEric

Okay! Yeah, applying is usually the tricky part, because you have to go through all of your knowledge and think about which tools to apply. So, while the person pushes the trampoline down 8cm, he or she is doing work on the trampoline. That work is simply stored as spring potential energy. That's what happens whenever you compress a spring. Is that okay with you?

13. anonymous

Kind of, yes.

14. theEric

Okay! Well, mechanical energy is KE and PE. This person is adding mechanical energy to the system with his or her muscle stuff. That energy is used to stretch the trampoline, so you can say that that energy goes to stretching the trampoline. That energy is stored in the trampoline, now, as potential energy. It has been displaced by some amount $$x$$, so it now has the potential energy $${\rm PE}=\frac12kx^2$$

15. theEric

How much energy is this really? It's the energy that the person does by exerting a force on the trampoline for 8cm. For a semi-complicated reason, we have to think about the average force. $$W=\overline F\dot\ d$$ Because of how the trampoline (and, ideally, springs) work, the work done on the spring is the energy that the spring has $$\rm that\ has\ potential\ to\ do\ work$$, which defines it as potential energy. If that's okay, then we recap. The work done on the trampoline is the potential energy of the trampoline.

16. theEric

So, mathematically, $$W=\rm PE$$ From that, we're fortunate to see that $$\overline F\dot\ d=\frac12kx^2$$ Which is fortunate, because we can find the average force. If you have a question about why we need to look at the average force, feel free to ask.

17. anonymous

Okay, so I think I understand this now. So to find the force applied by the person's legs we are trying to find the "F" in the equation $F.d = 0.5k x^{2}$ the 'd' in the equation will represent the distance that the person moved which is the 8cm. and the 'x' in the question will be the total distance traveled by the spring (right..?) so 8cm + the answer gained from question 1. Is that right...?

18. theEric

Close! You've got it most of the way! But this doesn't involve problem one! You left off at $$\overline F\dot\ d = 0.5k x^{2}$$ and said that $$d=x=8\rm cm$$ What you can do is either plug in $$d=x=8\rm cm$$, or you can plug in just $$d=x$$. I'll plug in just $$d=x$$ just so the equation is left general. Just because. $$\overline F\dot\ x = 0.5k x^{2}$$ From there, you can divide by $$x$$ to get $$\overline F = 0.5k x$$ As @thushananth01 said, you now have all the info you need to find that average force :)

19. theEric

How is that? Any questions?

20. anonymous

oh okay, I thought that you needed to consider the energy that was stored before the trampoline was pushed an extra 8cm (because the person was standing it, then pushed it...I think..) and my question is this: so for questions like these we don't have to consider the gravitational potential at all?

21. theEric

You, just in that paragraph, asked two very good questions! I have to go, but I will answer these when I get back. Briefly, we want to consider that the work done by the person contributes to a certain change in the potential energy, one that we know. We might have been able to use gravitational PE for the first problem, but I have to think about it.

22. anonymous

oh okay, thank you so much.

23. theEric

You're welcome! Let me know if you have other questions! Take care!