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Spring Elastic Potential Energy

by Icetray
Tags: elastic, energy, potential, solved, spring
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Icetray
#1
Jan1-08, 08:15 AM
P: 85
Hi,

I have a question (link below with scanned page) on a spring, some load and I have the strain energy of the spring.

Can anyone help me double check (a) and guide me along with (b) and (c)? From there I think I know how to do part (d).

Your help is much appreciated.

Link for question:
http://img81.imageshack.us/img81/316...ment002rx3.jpg
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Oerg
#2
Jan1-08, 08:43 AM
P: 364
Hi Icetray

For part (a), your answer is correct, have confidence in your answer!

part (b) do you notice that at equilibrium, the force exerted by the spring is equal to the weight of the mass? Equate both equations together to find out the spring constant. Then you can use the equation for the elastic potential energy. You should not have equated the gravitational potential energy loss with the spring potential energy equation, because they are not equal! we shall discuss this later.

part (c) Where does the gravitational potential energy go to? Part of it goes to the elastic potential energy of the spring, then the other? Remember that when you put the load on the spring the spring starts to udnergo SHM, that means that at is equilibrium position, it has half K.E. and half P.E..We consider this mathematically

The work done that comes from increasing the potential energy of the spring to equilibrium is [tex] \int F dx=\int kx dx= \frac{1}{2}kx^2=\frac{1}{2}ke^2 [/tex]
The work done that goes into increasing the kinetic energy is thus the loss of gravitational potential minus the gain of elastic potential:
[tex] mgh-\frac{1}{2}ke^2 =ke^2-\frac{1}{2}ke^2=\frac{1}{2}ke^2[/tex]

After damping, the load comes to a rest and that means its K.E. has dissipated.

part (d) Now that you know elastic potential released by the spring, calculate the temperature change!
Icetray
#3
Jan1-08, 09:20 AM
P: 85
Quote Quote by Oerg View Post
Hi Icetray

For part (a), your answer is correct, have confidence in your answer!

part (b) do you notice that at equilibrium, the force exerted by the spring is equal to the weight of the mass? Equate both equations together to find out the spring constant. Then you can use the equation for the elastic potential energy. You should not have equated the gravitational potential energy loss with the spring potential energy equation, because they are not equal! we shall discuss this later.
Can you explain on this? Equate what equations together?

Oh yes, I will have more confidence. ;-)

Oerg
#4
Jan1-08, 09:58 AM
P: 364
Spring Elastic Potential Energy

im sure you have learnt

[tex]F=kx[/tex]this means that the elastic force rpesented by the spring when it is strentched for a displacement of x meters. At equilibrium, the force exerted by the spring is
equals to the weight of the mass. [tex] mg[/tex]

I used x for displacement and e for the displacement at equilibrium.
Icetray
#5
Jan1-08, 10:01 AM
P: 85
Quote Quote by Oerg View Post
im sure you have learnt

[tex]F=kx[/tex]this means that the elastic force rpesented by the spring when it is strentched for a displacement of x meters. At equilibrium, the force exerted by the spring is
equals to the weight of the mass. [tex] mg[/tex]

I used x for displacement and e for the displacement at equilibrium.
So F(0.04) = (8.0)(9.81).

F will be in newtons right? What do I do next? Sorry, I'm really hopeless at this.
Oerg
#6
Jan1-08, 10:03 AM
P: 364
you should equate the elastic force posed by the spring, ke, with the weight of the mass mg since these two forces cancel out each toher at equilibrium.

[tex] ke=mg [/tex]
Icetray
#7
Jan1-08, 10:12 AM
P: 85
Quote Quote by Oerg View Post
you should equate the elastic force posed by the spring, ke, with the weight of the mass mg since these two forces cancel out each toher at equilibrium.

[tex] ke=mg [/tex]
Ok, so from that I obtain a value for k. Wit that value I use Fx? What is F in this case? I'm damn lost at the moment.
Oerg
#8
Jan1-08, 10:17 AM
P: 364
Hi Icetray

F is force. F=kx gives us the restoring force when you stretch a spring. At equilibium, the restoring force is equals to the weight of the load. Draw a freebody diagram if you are unsure. The restoring force acts towards the top while the weight acts downwards.

With k, you can now use the equation for the elastic potential

[tex] P.E.=\frac{1}{2}ke^2[/tex]

e stands for the displacement at equilibrium. At equilibrium, x=e.
Icetray
#9
Jan1-08, 10:21 AM
P: 85
Quote Quote by Oerg View Post
Hi Icetray

F is force. F=kx gives us the restoring force when you stretch a spring. At equilibium, the restoring force is equals to the weight of the load. Draw a freebody diagram if you are unsure. The restoring force acts towards the top while the weight acts downwards.

With k, you can now use the equation for the elastic potential

[tex] P.E.=\frac{1}{2}ke^2[/tex]

e stands for the displacement at equilibrium. At equilibrium, x=e.
Wouldn't the EPE just be equal to the mgh answer than? From your equation, I'd be finding the value of k. Who does this help me find the EPE?
Oerg
#10
Jan1-08, 10:22 AM
P: 364
im sorry i should ahve been more specific when i put down P.E., the P.E. that you quoted refers to the elastic potential energy
Icetray
#11
Jan1-08, 10:43 AM
P: 85
Thanks! I've finally solved everything! You rock Oerg!


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