Using Conservation of Energy to Find Friction Coefficient

AI Thread Summary
A 2.00 kg block on a rough incline connected to a spring is analyzed to find the coefficient of kinetic friction after it moves 0.204 m down the incline. Initial attempts to calculate the friction coefficient using energy conservation yielded a negative value, indicating an error in approach. Suggestions include using total mechanical energy before and after the block's release, factoring in energy lost to friction. The correct method involves equating gravitational potential energy to spring potential energy and work done against friction. Clarifications are made regarding the correct application of the cosine of the incline angle in the work-energy calculations, emphasizing the need to accurately represent the forces involved.
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Homework Statement


A 2.00 kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of 100 N/m. The block is released from rest when the spring is unstretched, and the pulley is frictionless. The block moves 0.204 m down the incline before coming to rest. Find the coefficient of kinetic friction between the block and incline.
http://img692.imageshack.us/img692/9474/physicsam.jpg

Homework Equations





The Attempt at a Solution


Final position is used as 0 height.
http://img694.imageshack.us/img694/2420/physicsm.jpg
The way I see the forces is that Fs and Ffric acts in one direction and Fgsin37 opposes it.
Fs+Ffric=Fgsin37
kx+u(mgcos37)=mgsin37
100(0.204)+u((2)(9.8)cos37)=(2)(9.8)sin37
u=-0.55
Coefficient came out negative so this approach won't work.

For the most part, I don't understand how to approach this problem. What does it seem like I don't understand?
 
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312213 said:

Homework Statement


A 2.00 kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of 100 N/m. The block is released from rest when the spring is unstretched, and the pulley is frictionless. The block moves 0.204 m down the incline before coming to rest. Find the coefficient of kinetic friction between the block and incline.
http://img692.imageshack.us/img692/9474/physicsam.jpg

Homework Equations





The Attempt at a Solution


Final position is used as 0 height.
http://img694.imageshack.us/img694/2420/physicsm.jpg
The way I see the forces is that Fs and Ffric acts in one direction and Fgsin37 opposes it.
Fs+Ffric=Fgsin37
kx+u(mgcos37)=mgsin37
100(0.204)+u((2)(9.8)cos37)=(2)(9.8)sin37
u=-0.55
Coefficient came out negative so this approach won't work.

For the most part, I don't understand how to approach this problem. What does it seem like I don't understand?

If you are going to use the conservation of energy why don't you use all the mechanical energy you have before the mass is released. and set that equal to all the mechanical energy after plus the "loss" of mechanical engergy due to friction as thermal energy.

If the plane was frictionless the mass would be displaced further down the plane as one way to view it, yes? Then it seems to me you can figure out the normal force on the block. And the force of static friction can be figured using the energy/work. Then you have mu? This is a cursory observation of the situation that I believe will work.

I think the title of your question said using the conservation of energy? So maybe you start off with all GPE to begin with and end with spring pot. energy 1/2k(x^2) and heat(due to friction)... does this stuff strike you as familiar?
 
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Ui=Kf+Usf+Ef
mgh=0+1/2kx²+E
(2)(9.8)(0.204sin37)=(1/2)(100)(0.204)²+E
0.3255J=E

The nonconserved energy is the work of friction so
W=(Ffric)rcos37
0.3255=u((2)(9.8)cos37)(0.204)cos37
u=0.128
(0.12763323293335829568080338705843)

Does the nonconserved energy convert to mu like that?
 
That looks good. BTW, what the two different results suggests is that your assumption about the direction of the frictional force at the new equilibrium is wrong. In other words , it is working to oppose the block being pulled back up the incline. That the magnitude of 0.55 is also wrong suggests that the static and kinetic cooefients are different--since you were asked to find the kinetic coefficient, your second approach was the right one.
 
The work seems to have worked correctly up until the part of
W=(Ffric)rcos37
0.3255=u((2)(9.8)cos37)(0.204)cos37

One of those cos37 shouldn't be there. Which is it? Or are either acceptable to be there, so long as only one is?
 
312213 said:
The work seems to have worked correctly up until the part of
W=(Ffric)rcos37
0.3255=u((2)(9.8)cos37)(0.204)cos37

One of those cos37 shouldn't be there. Which is it? Or are either acceptable to be there, so long as only one is?
That last cos 37 is not correct. W_f = F_f . d = uN(d)cos alpha, where alpha is the angle between the friction force and displacement vectors, alpha = ?
 
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