# Kinetic Friction of block on table

1. Oct 8, 2007

### Heat

1. The problem statement, all variables and given/known data
Consider the system shown in the figure. Block A has weight w_A and block B has weight w_B Once block B is set into downward motion, it descends at a constant speed.

Calculate the coefficient of kinetic friction between block A and the table top.

A cat, also of weight w_A, falls asleep on top of block A. If block B is now set into downward motion, what is the magnitude of its acceleration?

2. Relevant equations

F=ma
w=ma

3. The attempt at a solution
I know that when the box is sliding at constant speed the kinetic friction is: Fsubk = $$\mu$$subk n .

The thing is I don't know where to start, as there are no numbers.

I would know that the block b is gravity pulling down.

2. Oct 8, 2007

### Heat

ok, I just read that an object is at equilibrium whether at rest or constant velocity.

when the crate moves, the forces would be:

Sum Fx = T - fk =0
Sum Fy = n-w = 0

Last edited: Oct 8, 2007
3. Oct 8, 2007

### Staff: Mentor

Do the first part first. (Duh!) Just analyze the forces and apply Newton's 2nd law to each mass separately, as usual. Take advantage of the fact that it moves at constant speed. What does that tell you?

4. Oct 8, 2007

### Heat

Well I edited my previous post,

but I found that fk = T and n=w

using fk = usubkn

uk = fk/n

5. Oct 8, 2007

Good:

6. Oct 8, 2007

### Heat

so the coefficient is fk/n, for part one?

The "Express your answer in terms of w_a,w_b,g " is for the second question. :)

I just placed fk/n as the answer, and I good feedback saying "The correct answer does not depend on the variables: f_k, n." :(

ok, I got it, indeed with using w as the variable. :)

Last edited: Oct 8, 2007
7. Nov 2, 2010

### malakian_77

1.) µK = Wb/Wa

2.) A cat, also of weight w_A, falls asleep on top of block A

a = [Wa*g*(Wb/Wa)] / [2*Wa + Wb ]

since the cat has a weight of A, and the block has Weight A
the above equation needs to be (2*Wa) where Wa is present.