# Work of kinetic friction on block in vertical circle

1. Feb 12, 2012

### galaticman

1. The problem statement, all variables and given/known data

1. The problem statement, all variables and given/known data
A block of mass 0.015kg enters the bottom of a circular, vertical track with a radius R = 0.3m at an initial velocity of 4/ms. If the block loses contact with the track at an angle of 130 degrees, what is Wk, the work done by kinetic friction?

2. Relevant equations

F=ma
Wtot= Wmg+WN+WFk=1/2*mvf2-1/2*mvi2

3. The attempt at a solution

So far I have that Wmg=-mg*(Rcos50+R)=-0.072447J and that WN=0 J
The only part that I am having trouble with is finding vf right before that block leaves the track. Any help would be appreciated!!

EDIT: Im not sure how to go about relating F=ma=mv2/R without knowing what the friction force is ie. fk

Last edited: Feb 12, 2012
2. Feb 12, 2012

### PeterO

3. Feb 12, 2012

### PhanthomJay

When the block loses contact with the track, the normal force is 0. So only the radial component of the weight force contributes to the centripetal acceleration at that point.

4. Feb 12, 2012

### galaticman

So if I am understanding this correctly would that mean that:
1. Fxmg=mv2/R
2. mg*sin40=mv2/R
3. vf=√(R*9.8*sin40)
4. vf=1.3747m/s

AND would that also mean that the mass of the block did not matter for Vf

Last edited: Feb 12, 2012
5. Feb 12, 2012

### PhanthomJay

yes
Well, it cancels out, but if the mass was higher, the block would not travel 130 degrees, it would travel less since there is more friction force .. So it does matter, even though you don't need to know it to solve this problem.

I see that per PeterO, help is already being provided in a separate post from another OP. I might just add that instead of using $W_t = \Delta KE$,
which is OK to use, you might want to use the alternate form of the Work-Energy Theorem , $W_{nc} = \Delta PE + \Delta KE$,
where since the normal force does no work and since gravity is not a non-conservative force, then $W_{nc} = W_{friction}$

6. Feb 12, 2012

### galaticman

thanks so much! I understand the entire problem completely now! you were a huge help