AP Physics 1 Help? Centripetal motion/Kinematics/Friction problem

[tortilla]

Homework Statement

A track consists of frictionless arc XY which is a quarter circle of radius R and a rough horizontal secion YZ. Block A of mass M is released from rest at point X slides down the curved section of the track to point Y. Block A continues moving along the horizontal track to the right, sliding past point P which is a distance l from point Y. The coefficient of kinetic friction between the blocks and the horizontal part of the track is mu.

Express your answers in terms of M, l, mu, R, and g.

1. Determine the speed of block A at point Y.

2. Determine the force due to friction acting upon block A as it slides horizontally.

3. Determine the work done by friction on block A as it passes point P.

4. Determine the speed of block A as it passes point P.

Relevant Equations

Fc = (Mv^2)/R
Ff = Fn*mu
w = Ff*d

I think I have solved the first three, and only really need help on question four.

For number one, I used Fc = (Mv^2)/R and just rearranged it for velocity so I ended up with v = sqrt(ac * R)

For number 2 I used Ff = Fn*mu and got Mg*mu = Ff

For number 3 I used w = Ff*d and got w = -Mg*mu*l

For number four I have no idea.

 

[osilmag]

The potential energy should equal the kinetic energy minus the work of friction. That should be what you use to find the velocity.

 

[PeroK]

Homework Statement:: A track consists of frictionless arc XY which is a quarter circle of radius R and a rough horizontal secion YZ. Block A of mass M is released from rest at point X slides down the curved section of the track to point Y. Block A continues moving along the horizontal track to the right, sliding past point P which is a distance l from point Y. The coefficient of kinetic friction between the blocks and the horizontal part of the track is mu.

Express your answers in terms of M, l, mu, R, and g.

1. Determine the speed of block A at point Y.

2. Determine the force due to friction acting upon block A as it slides horizontally.

3. Determine the work done by friction on block A as it passes point P.

4. Determine the speed of block A as it passes point P.
Homework Equations:: Fc = (Mv^2)/R
Ff = Fn*mu
w = Ff*d

I think I have solved the first three, and only really need help on question four.

For number one, I used Fc = (Mv^2)/R and just rearranged it for velocity so I ended up with v = sqrt(ac * R)

For number four I have no idea.

Your answer to number 1 is not correct. What value are you going to use for $a_c$? You've used an equation for the centripetal acceleration associated with constant speed circular motion, which is not relevant here.

If you work out how to solve part 1, you may see how to solve part 4.

 

[tortilla]

Your answer to number 1 is not correct. What value are you going to use for $a_c$? You've used an equation for the centripetal acceleration associated with constant speed circular motion, which is not relevant here.

If you work out how to solve part 1, you may see how to solve part 4.

Thanks so much for pointing that out! should I be using v=(2piR)/t? I don't know the time it was on the track

 

[PeroK]

Thanks so much for pointing that out! should I be using v=(2piR)/t? I don't know the time it was on the track

This problem will be difficult using forces and acceleration. Can you think of another approach?

 

[tortilla]

I'm really not sure. Should I be looking at the angular speed? just v/R?

 

[PeroK]

I'm really not sure. Should I be looking at the angular speed? just v/R?

No. Something else. You used it to solve part 3.

 

[tortilla]

No. Something else. You used it to solve part 3.

Because it isn't uniform the work is not zero?

 

[tortilla]

No. Something else. You used it to solve part 3.

since at point X it is at rest all the energy is gravitational potential, right? So if I set mgR equal to .5*m*v^2 and get v = sqrt(2gR) is that correct for part 1?

 

[haruspex]

since at point X it is at rest all the energy is gravitational potential, right? So if I set mgR equal to .5*m*v^2 and get v = sqrt(2gR) is that correct for part 1?

Yes.

 

[tortilla]

Yes.

Thank you so so much