Question

34. A 40kg slab rests on a frictionless floor. A 10kg
block rests on top of the slab. The coefficient
kinetic friction between the block and the slab
is 0.40. A horizontal force of 100N is applied
on the 10kg block. Find the resulting
acceleration of the slab. (g = 10ms-)
100 N + 10kg block
40kg slab​

Answer 1

Let us first analyse if these blocks move together.

For that we will find the acceleration of the total mass (40+10) kg first.

Using formula F=ma or a= mF = 50kg.100N

.

The acceleration is 2m/s 2

and the maximum frictional force =μ s ×N

=μ s ×m 1 g

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

Thus, we see 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=39.2N

Resulting acceleration of the slab , a=F/m=39.2/40=0.98m/s 2

Thus, the resulting acceleration of the slab is 0.98m/s 2solution

Answered By=shreyash. ...