Using Conservation of Energy to Find Friction Coefficient

In summary: W_f = uN(d) = uN(d)cos alpha = uN(d)1 = uN(d) = u(mg . cos alpha)(d) = u(mg . cos 0)(d) = u(mg)(d) = u(mgd) = u(mgh) = u(mg)(d . cos 0) = u(mg)(d)(1) = u(mg)(d) = u(mgd) = u(mg)(d) = u(mg)(d) = u(mgh), where d = h = mgh/mg = h = u(mg)/mg = u(mg) = u(m
  • #1
312213
52
0

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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  • #2
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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  • #3
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?
 
  • #4
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.
 
  • #5
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?
 
  • #6
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 = ?
 

Related to Using Conservation of Energy to Find Friction Coefficient

1. What is conservation of energy and how does it relate to finding friction coefficient?

Conservation of energy is the principle that energy cannot be created or destroyed, only transferred or transformed. In the context of finding friction coefficient, this means that the total amount of energy in a system before and after a frictional force acts on it should remain constant. This allows us to use the initial and final energy values to calculate the amount of energy lost to friction and therefore determine the coefficient of friction.

2. What are the different methods for using conservation of energy to find friction coefficient?

There are several methods for using conservation of energy to find friction coefficient, including the inclined plane method, the Atwood's machine method, and the pendulum method. Each method involves measuring the initial and final energy values of a system and using these values to determine the amount of energy lost to friction.

3. How accurate is using conservation of energy to find friction coefficient?

The accuracy of using conservation of energy to find friction coefficient depends on several factors, such as the precision of the measurements taken and the assumption that all other forces (such as air resistance) are negligible. In general, this method can provide a good estimation of the coefficient of friction, but it may not be completely accurate.

4. Can conservation of energy be used to find friction coefficient in all situations?

No, conservation of energy cannot be used to find friction coefficient in all situations. This method is most accurate when the frictional force is the only significant force acting on the system. In more complex situations where there are multiple forces at play, other methods such as Newton's laws of motion may be more appropriate.

5. How can the coefficient of friction be used in practical applications?

The coefficient of friction is an important quantity in many practical applications, such as designing brakes for vehicles or determining the maximum weight a surface can support. Knowing the coefficient of friction allows us to predict and control the effects of friction in various systems, making it a crucial factor in engineering and everyday life.

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