Finding the coefficient of kinetic friction and work

[leafeater47]

In a truck loading station at a post office, a small 2.00-kg package is released from rest at point A on a track that is one-quarter of a circle with radius 1.60 m. The size of the package is much less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest.

a) What is the coefficient of kinetic friction on the horizontal surface?

b) How much work is done on the package by friction as it slide down the circular arc from A to B?Normally, I would follow the guidelines and post equations and attempts at this problem but the thing is, I have no idea what equations I'm supposed to use, and that's a bit of a problem.

*EDIT* Equations: Fy = -mgy + N
I plugged everything in, getting 0 = -(2)(9.8)(cos0) + N and N=19.6. Is this right? What do I do from here, if it is right?

 

[alphysicist]

Hi leafeater47,

In a truck loading station at a post office, a small 2.00-kg package is released from rest at point A on a track that is one-quarter of a circle with radius 1.60 m. The size of the package is much less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest.

a) What is the coefficient of kinetic friction on the horizontal surface?

b) How much work is done on the package by friction as it slide down the circular arc from A to B?


Normally, I would follow the guidelines and post equations and attempts at this problem but the thing is, I have no idea what equations I'm supposed to use, and that's a bit of a problem.

*EDIT* Equations: Fy = -mgy + N
I plugged everything in, getting 0 = -(2)(9.8)(cos0) + N and N=19.6. Is this right? What do I do from here, if it is right?

That looks like the right value for the normal force (for the horizontal surface). How is the normal force related to the kinetic frictional force?

The effect of friction is to bring the object to rest in a distance of 3m (starting at 4.8m/s). What can you find from those values, that will relate to the frictional force?

 

[leafeater47]

So, I should use fK = coeff. of friction x N?
To get fK, do I do 4.8m/s x 3m x .2kg? That doesn't seem right...

OR, when you said what can i find from the values 4.8m/s and 3m, did you mean find the time for how long it takes the block to stop sliding? If that's what you meant I got .625s.

And if that's the case, then to get fK i would do (4.8m/s)(.2kg)(.625s) and get .6 and when I plug that into the equation above, I get coeff. of friction = .031.

Is any of this right or am I just completely spouting nonsense? I don't think I did anything right lol I think the coeff. of friction is supposed to be bigger.