Q.
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 of friction is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If $g = 9.8\ ms^{- 2}$, the resulting acceleration of the slab will be
A) 1.47 $ms^{- 2}$
B) 1.69$\ ms^{- 2}$
C) 9.8$\ ms^{- 2}$
D) 0.98$\ ms^{- 2}$
Explanation
(d) $f_{ms} = 0.6 \times 10 \times 9.8$ N
$= 58.8\ $N
Since the applied force is greater than $f_{ms}\ $therefore the block will be in motion. So, we should consider $f_{k}$
$f_{k} = 0.4 \times 10 \times 9.8$ N or $f_{k} = 4 \times 9.8$ N
This would cause acceleration of 40 kg block
Acceleration = $\frac{4 \times 9.8\ N}{40\ kg} = 0.98\ ms^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.