Question

A 5.00 kg block rests on a level table. The coefficient of friction between the block and the table is 0.55. A 4.00 kg mass is attached to the block by a string of negligible mass passed over a light frictionless pulley (as shown below). What is the acceleration of the block when the 4.00 kg weight is released?

Answer 1

5.00 kg 4.00 kg

There are 4 forces acting on the resting block. First, gravity pulling it down. Next, the normal force pushing the resting block up. Third, the gravitational force from the other block. Last, the friction force from the surface.

Apply Newton's laws of motion. F=MA.

The force of gravity must equal the normal force (or else it would be floating or sinking).

Normal = Gravity = M*g = 5 * 9.8 = 49 Newtons.

The frictional force is dependent upon the normal force (not weight force). Friction = $\mu$*N = 0.55*49 = 26.95N.

The weight of the second block is W = M*g = 4*9.8 = 39.2 Newtons.

m₂g+m₂a=$\mu$*N-m₁a

=a(m₂+m₁)=39.2-26.95=12.25

a=12.25/9=1.4m/s²

So acceleration is a=1.4m/s²