Q. A $40\ kg$ slab rests on a frictionless floor as shown in the figure. 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 $100\ N$. If $g = 9.8\ m/s^{2}$, the resultant acceleration of the slab will be
A) $0.98\ m/s^{2}$
B) $1.47\ m/s^{2}$
C) $1.52\ m/s^{2}$
D) $6.1\ m/s^{2}$
Explanation
(a) Limiting friction between block and slab $= \mu_{s}m_{A}g$
$= 0.6 \times 10 \times 9.8 = 58.8\ N$
But applied force on block $A$ is $100\ N$. So the block will slip over a slab
Now kinetic friction works between block and slab
$F_{k} = \mu_{k}m_{A}g = 0.4 \times 10 \times 9.8 = 39.2\ N$
This kinetic friction helps to move the slab
$\therefore$ Acceleration of slab $= \frac{39.2}{m_{B}} = \frac{39.2}{40}$
$= 0.98\ m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.