## Spring Problem (Urgent)

To be honest I don't what to do........

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 Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus Saying that you have no idea doesn't count as an attempted solution. Have you read your class notes and/or text?
 my professor doesn't go in details and book is to consusing but still I will give a try: attempt: First I thought to use W = delta KE + delta PE but I was confused b/c i did not see a way to put those spring forces in there so i was kindly asking for a help

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## Spring Problem (Urgent)

 Quote by dsptl my professor doesn't go in details and book is to consusing but still I will give a try: attempt: First I thought to use W = delta KE + delta PE but I was confused b/c i did not see a way to put those spring forces in there so i was kindly asking for a help
Thank you for trying, your threads will generally get answered a lot quick if you show an attempted solution.

You're right to approach the problem using conservation of energy. You know that the work done by friction and the 400 N/m spring must be equal to the energy stored in the N/m spring.

Can you go from here?

 is it like this: F x S = Ef - Ei

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 Quote by dsptl is it like this: F x S = Ef - Ei
We'll take it a step at a time. What is the energy stored in the 200N/m spring before the block is released?

 = .5k(x.x) + .5mv.v + mgh

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 Quote by dsptl = .5k(x.x) + .5mv.v + mgh
Could you explain how each of the terms arise, what does each term represent?

 1st is spring force , 2nd is kinetic energy and 3rd is potential enery, which is 0.

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 Quote by dsptl 1st is spring force , 2nd is kinetic energy and 3rd is potential enery, which is 0.
Good. What about the kinetic energy, is that non-zero before the block is released?

 i think it is zero since Vi is 0

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 Quote by dsptl i think it is zero since Vi is 0
Correct. So, the total energy of the system is simply the energy stored by the spring, can you calculate it?

Now, when the block reaches the 400N/m spring is compresses it. So how much energy is stored in this spring? Can you also calculate this value?

 at intial = .5(200N/m)(.1 x .1) = 1N at Final = .5(4000N/m)(.05 x .05) = .5 N
 do i just add them together?

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 Quote by dsptl do i just add them together?
No.

You started off with this much energy:
 Quote by dsptl at intial = .5(200N/m)(.1 x .1) = 1N
And ended up with this much energy:
 Quote by dsptl at Final = .5(4000N/m)(.05 x .05) = .5 N
Where did the rest of the energy go?

You also need to be careful with your units, the Newton is not a measure of energy.