Question

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The coefficient of friction between the block and the slab is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 10 $m/s^2$ the resulting acceleration of the slab will be:

• A. 2.0 $m/s^2$
• B. 1.47 $m/s^2$
• C. 1.52 $m/s^2$
• D. 6.1 $m/s^2$

Answer 1

The correct option is A 2.0 $m/s^2$

Given: a 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The coefficient of friction between the block and the slab is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N.

To find the resulting acceleration of the slab if $g=10m/s^2$

Solution:

In present case maximum static frictional force

$=\mu mg⟹=0.4\times 10\times 100⟹=400N$

When a force of 100N is applied on 10kg there would not be relative motion between the slab and block so acceleration of both would be same.

$m\times a=F⟹(10+40)\times a=100⟹a=2m/s^2$

Hence the resulting acceleration of the slab $2m/s^2$