Solving Kinetic Friction PB: How Many Times Does Block Cross Rough Section?

[Footballer010]

Homework Statement

A block of mass m is initially held in place on the left side of a track at a distance h above the bottom of the track. The track is completely frictionless, with the exception of a rough horizontal section of track of length h which has a coefficient of kinetic friction μ = 0.11. If the block is released from rest, how many times does the block completely cross the rough section of track before it stops?

a. 6 times
b.9 times
c. 11 times

Homework Equations

U=mgh
W=\frac{1}{2}mv_f²-\frac{1}{2}mv_i²

The Attempt at a Solution

I know that the potential energy at the top is mgh, but other than that I'm lost. Could use some help.

 

[Bacat]

No work is done on the block while the block is on the frictionless part of the track. So you can ignore these parts of the track. You can calculate how much work is done on the block as it passes through the friction patch from the work-energy theorem. Once you know how much energy is lost crossing the patch, you can figure out how many times the block could pass through the friction patch before it has no more energy. In the end, the block will be stuck in the friction patch but you can calculate how many times it went through it.

Make sense?

 

[Footballer010]

ok thanks, I got 9, but how would I set up the problem?

mgh=xu_kmgh?

where x equals the number of times the block crosses the section

that right?

 

[Bacat]

Well, the amount of work done on the block is defined by the work-energy theorem. I believe it is:w = \int_{b}^{a} f * d where f is the force applied and d is the distance the force is applied through.

You know the only force doing work is the frictional force so if you know the length of the rough section you can calculate how much work is done by that section numerically.

Then you use the work-energy theorem (http://en.wikipedia.org/wiki/Work_(physics )) to show that this work is subtracted from the kinetic energy of the block. Since the block has no potential energy left when it is at the bottom of the track (it has all been converted to kinetic energy at that point) then you can think of the friction as being subtracted from the potential energy. Therefore, you can just divide the potential energy by the amount of energy (work) to cross the patch once and it will tell you how many times the block can pass. Since you know that the block must stop in the patch (it will never stop on a frictionless track), you will have a fractional value. The number of times it completely passed through the friction patch is the largest whole number smaller than that number.

 

[Footballer010]

The fractional value is \frac{100}{11}, but from the answers choices... b. 9 times is the closest.

Thanks for the help.

 

[Bacat]

\frac{100}{11} = 9.09 so you can see that the block crosses 9 whole times before it runs completely out of energy. Actually, if you wanted to you could calculate how far the block can go into the friction path before it stops since you know it is .09 times the potential energy.