Inclined plane and spring problem

[emailanmol]

http://imageshack.us/photo/my-images/194/inclinedplane.png/



1.Homework Statement

The block is released from rest with the spring in the
unstretched position. The block moves 10 cm down the incline before coming to rest.
Find the coefficient of friction between the block and the incline. Assume that the
spring has a negligible mass and the pulley is frictionless.

Go to the link below for the figure

http://imageshack.us/photo/my-images/194/inclinedplane.png/

The Attempt at a Solution

I solved it using energy concept.

ie work done by spring force + work done by friction + work done by gravity = 0(no change in KE)

ie
-1/2*k*(x^2) - u*mg*cos(37)*x + mgsin(37)*x=0

that leads to

1/2*k*x=mg{sin 37 -cos(37)*u)-------Remember this as equation (1)

I plug in and get the answer as nearly 0.125 :-)

-------------------------------------------------Now the real problem

What if i use force equation.
At the point the block comes to rest, three forces act on the block along the inclined plane namely friction(acts upward), spring force(acts upward) and weight component along the plane(acts downward)

so kx + umgcos(37)=mgsin(37)

ie kx=mg{sin(37)-u*cos(37)}

which contradicts equation (1) i got earlier as the LHS there was K*x*1/2What mistake am i making?
Please help

 

[omoplata]

OK, I think I got it, but I'm not sure.

I think your first solution using the conservation of energy is correct.

But the second solution using Newton's 2nd law is not. You assumed that at equilibrium the frictional force acts upward along the string and is $F_{friction} = \mu m g \cos(37^{\circ})$. But actually the static frictional force can be anywhere within the range $\mu m g \cos(37^{\circ}) \geq F_{friction} \geq - \mu m g \cos(37^{\circ})$.

 

[emailanmol]

Hmm.I don't think this reasoning is correct.

Lets take the instant just before the block stops. The extension in spring is nearly 10 cm.Gravity component is constant and kinetic friction is acting as a very small magnitude of velocity is still there.Also acceleration is tending towards zero.

Since kinetic friction is acting it is N*u where N is our normal reaction constant at Mg*cos(37).

That gives our equation as nearly the force equation back.
So there has to be another reason.

 

[omoplata]

Lets take the instant just before the block stops. The extension in spring is nearly 10 cm.Gravity component is constant and kinetic friction is acting as a very small magnitude of velocity is still there.Also acceleration is tending towards zero.

I agree with everything here except the claim that the acceleration is tending towards zero. It is actually decelerating at the higest rate. The component of the net force on the block in the upward direction along the string is $F_{1} = k x + \mu m g \cos(37^{\circ}) - m g \sin(37^{\circ})$. This keeps increasing as $x$ increases. The block initially started moving with some acceleration in the downward direction along the string. at that moment $F_{1} < 0$. At some point during its downward movement the block attained zero acceleration when $F_{1} = 0$. That's when it had the highest velocity. Then as $x$ increased $F_{1}$ increased so that $F_{1} > 0$, and the block started decelerating in the downward direction (i.e. started accelerating in the upward direction). At the moment just before the block stops $x$ is at a maximum, therefore $F_{1}$ is at a maximum and so the deceleration is at a maximum. If you apply Newton's 2nd law for that moment you need to take that deceleration into account.

Then the block attained zero velocity and started moving in the upward direction. At that moment the frictional force changed and started pointing in the downward direction (The dynamic friction became static friction. The static friction can take any value between $\mu m g \cos(37^{\circ}) \geq F_{friction} \geq - \mu m g \cos(37^{\circ})$). If its magnitude was not enough to completely stop the upward acceleration when $F_{friction} = - \mu m g \cos(37^{\circ})$, the block would undergo a damped harmonic motion. But since they say the block stops we have to assume that it was enough to completely stop the upward acceleration. So the acceleration is zero and the velocity is zero and the block comes to a complete standstill. The final static frictional force is not $\mu m g \cos (37^{\circ})$. It gets settled somewhere between $\mu m g \cos (37^{\circ})$ and $- \mu m g \cos (37^{\circ})$ so that the acceleration is zero.

 

[emailanmol]

I got it Omoplata.
Thank you so much.You are dead right :-)