Question

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 $g=9.8m/s^2$ , then the resulting acceleration of the slab will be

• A. $0.98m/s^2$
• B. $1.47m/s^2$
• C. $1.52m/s^2$
• D. $6.1m/s^2$

Answer 1

Answer: (A)

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 $a=\frac{F}{m}=\frac{100N}{50kg}$.

The acceleration is $2m/s^2$ and the maximum frictional force $={\mu }_s\times N$

$={\mu }_s\times m_{1}g$

$=\left(0.60\right)\left(10\right)\left(9.8\right)=58.8N$

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 $F=\mu mg$

$=0.4\times 10\times 9.8=39.2N$
Resulting acceleration of the slab , $a=F/m=39.2/40=0.98m/s^2$

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