Question

A 47.0 kg slab rests on a frictionless floor. A 18.0 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.50, while the kinetic coefficient of friction is 0.40. The top block is acted upon by a force of 120 N. What is the acceleration (in meters/second^2) of the bottom block?

Answer 1

The block on top can support a pull of 0.50*18*g = 88.2 N without slipping. Since the applied force is greater than that, it will slip and there will be a friction force of
Ff = 0.4*18*g = 70.56 N between the blocks. That will help pull the lower block forward.
The lower block will accelerate at a rate
a = Ff/47 = 70.56/47
= 1.50 m/s^2