One block on another

1. Oct 7, 2009

Filip89

1. The problem statement, all variables and given/known data
A 104.0-kg block is on a horizontal frictionless surface, on top of it is a second smaller block of 68.0-kg. A horizontal force F of 613.4 N is applied to the upper block which accelerates with 5.84 m/s2. 1. Find the coefficient of kinetic friction between the two blocks.
2. Find the acceleration of the lower block while the two blocks are in contact.

2. Relevant equations
F=ma, fk=(coeff kinetic)* F_Normal

3. The attempt at a solution
I found the coefficient of kinetic friction by taking the horizontal force (613.4N) and subtracting from it the force needed to move the block which is F=ma (68.0*5.84). That gave me a fk of 216.28N which I then divided by the normal force (mg= 666.4) to get the coefficient.

I am having trouble finding the acceleration of the bottom block, I thought doing F=ma
(613.3-216.28)=(104+68)a would give me the acceleration but it doesn't. Can someone give me a hint as in what to do because I can't figure it out.

2. Oct 7, 2009

PhanthomJay

In part 1, you looked at the forces acting on the top block. In part 2, you need to draw a free body diagram of the bottom block, and examine the forces acting on the bottom block.

3. Oct 7, 2009

Filip89

Ok, so I add the two masses together (i assume) which gives me 172 kg all together. The horizontal force is still 613.4 N so i set up the equation F=ma 613.4N=(172kg)*a which then gives me an acceleration of 3.566 m/s^2 but that's incorrect.

4. Oct 9, 2009

PhanthomJay

No, when you take a free body diagram of the lower block, you 'isolate' it by itself and examine the forces acting on it. The 613.4 N force doesn't show up in such a diagram. What does show up is, amongst other forces, is the horizontal friction force from the top block. Use Newton 3 and 2 solve for the acceleration of the lower block. Make sure to use the correct value for the mass when using Newton 2. You shouldn't be adding the 2 masses together using this free body diagram approach.