A block (4.00-kg) is at rest on the right edge of a large 11.0-kg slab. There is

Question

A block (4.00-kg) is at rest on the right edge of a large 11.0-kg slab. There is no friction below the slab and the horizontal surface it is on but there is friction between the block and the slab. The coefficient of kinetic friction between the two is 0.25. The slab is 3.00 meters wide and a constant horizontal force is applied to the left on the block. If it takes 2.00 s for the block to reach the other side of the slab, (a) what is the magnitude of the force applied to the block? (b) How far does the bottom block travel? (c) What is the final velocity of the top and bottom block? (Hint: use a non-moving/accelerating reference frame, i.e. static origin on the right side of the table)

Explanation / Answer

The force of friction between the slab and block = 0.25(4)(9.8) = 9.8 N
This force is the force exerted on the slab.
The force on the block is F - 9.8
acceleration of the block = (F - 9.8)/4
acceleration of the slab = 9.8/11 = 0.89 m/s^2
Distance the slab moves call X
The distance the block moves = X + 3
Using the distance equation
0.5(0.89)(2)^2 = X
0.5(2)^2(F - 9.8)/4 = X + 3
X = 1.78
1.78 + 3 = 4.78 = (F - 9.8)/2
9.56 = F - 9.8
F = 19.36 N

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