Question

A $40kg$ slab rests on a frictionless floor. A $10kg$ block rests on top of the slab. The coefficient of kinetic friction between the block and slab is $0.4$. A horizontal force of $100N$ applied on $10kg$ block . Find the resulting acceleration of the slab.

Answer 1

Step 1: Given Data

Mass of the slab $m_s=40kg$

Mass of the block $m_b=10kg$

The coefficient of kinetic friction between the block and slab $\mu =0.4$

The horizontal force of $F=100N$ on $-x$ direction.

Step 2: Calculate the Common acceleration

The total mass of the system,

$m=m_s+m_b$

$\Rightarrow m=40+10=50kg$

Since acceleration is action on $-x$ direction, the pseudo force will be acting on $+x$ direction with a force of $50a$.

$\therefore 50a=100$

$\Rightarrow a=2m/s^2$

Step 3: Check acceleration of slab and block

Let us consider both the blocks move with the same acceleration.

Pseudo force on the slab $f_s=m_{s}a=40\times 2=80N$

Pseudo force on the block $f_b=m_{b}a=10\times 2=20N$

Again, the normal force $N=100N$

Therefore, limiting force, $f_L=\mu N=0.4\times 100=40N$

Since $f_s>f_L$, this is not the case here.

Step 4: Calculate resultant acceleration of slab

Therefore, the block and the slab do not move with the same acceleration.

If $a_s$ is the acceleration of the slab, then

$40a_s=40$

$a_s=1m/s^2$

Hence, the resulting acceleration of the slab is $1m/s^2$.