Question

A 40 kg slab rests on frictionless floor as shown in fig. A 10
kg block rests on the top of the slab. The static coefficient
of friction between the block and slab is 0.60 while the kinetic
friction is 0.40. The 10 kg block is acted upon by a horizontal
force of 100 N. If g = 9.8 m/s², the resulting acceleration of
the slab will be:
(a) 0.98 m/s² (b) 1.47 m/s²
(c) 1.52 m/s² (d) 6.1 m/s²

Answer 1

Explanation:

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=

m

F

=

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