A 40 kg slab rests on a frictionless floor a 10 kg block rests on

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60 while the kinetic coefficient is 0.40. The 10 kg block is acted upon by horizontal force 100 N. If g = 9.8m/s2, then what will be the resulting acceleration of the slab?

Friction Problem

Solution:

The total mass of the system is (10 kg + 40 kg = 50 kg).

Rearranging the formula F = ma, let us obtain the acceleration of the total mass as follows:

a = F/m

Substituting the values in the above equation, we get

a = 100 N/ 50 kg = 2 m/s2.

The maximum frictional force is given by the following formula:

= μs × N

Substituting the values in the above equation, we get:

= (0.60)(10) (9.8) = 58.8 N

From the above result, we notice that the frictional force is smaller than the applied force and proves that the block and the slab do not move relative to each other.

For the block, we have F = μmg

= 0.4 × 10 × 9.8 = 39.2 N

Resulting acceleration of the slab , a = F/m = 39.2/40 = 0.98m/s2

Thus, the resulting acceleration of the slab is 0.98m/s2

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