Question

A $40kg$ slab rests on a frictionless floor. A $10kg$ 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 of friction is $0.40$. The $10kg$ block is acted upon by a horizontal force of $100N$. The resulting acceleration of the slab will be

• A. $1.0m/s^2$
• B. $10m/s^2$
• C. $1.5m/s^2$
• D. $2.0m/s^2$

Answer 1

The correct option is A $1.0m/s^2$

Step 1: Find the maximum friction force between both the blocks.

Limiting static friction $f_l=\mu mg$ between both blocks,

$f_{max}=0.6\times 10\times 10N=60N$

Step 2: Find the friction force on the ground.

Since the applied force is greater than $f_{max}$ therefore the block will be in motion. So, we should consider$f_k.$

$f_k=0.4\times 10\times 10N=40N$

Step 3: Find the acceleration of the block.

Formula Used:

$a=\frac{F}{m}$

This would cause acceleration of $40kg$ block

Acceleration $=\frac{40}{40}=1.0m/s^2$

Final Answer: (D)