Question

A $40kg$ slab rests on a frictionless floor. A $10kg$ block rests on top of the slab (as shown in the figure). The coefficient of static friction ${\mu }_s$ between the block and slab is $0.60,$ whereas their kinetic friction coefficient ${\mu }_k$ is $0.40$. The $10kg$ block is pulled by a horizontal force $100.0N\hat{i}$. The resulting accelerations of the block and slab will be
(Take $g=10m/s^2$)

• A. $\left(1.0m/s^2\right)\hat{i},0$
• B. $\left(2.0m/s^2\right)\hat{i},-\left(2.0m/s^2\right)\hat{i}$
• C. $\left(6.0m/s^2\right)\hat{i},\left(1.0m/s^2\right)\hat{i}$
• D. $\left(4.0m/s^2\right)\hat{i},0$

Answer 1

The correct option is C $\left(6.0m/s^2\right)\hat{i},\left(1.0m/s^2\right)\hat{i}$

Mass of slab $=40Kg$

Mass of block$=10Kg$

${\mu }_s=0.6$

${\mu }_k=0.4$

So, static friction force

$={\mu }_{s}N$

$=0.6\times 10\times 10$

$=60N$

Kinetic friction force

$={\mu }_{k}N$

$=0.4\times 10\times 10$

$=40N$

In this case, the applied is $100N$which is greater than the maximum static friction force. So the block will move w.r.t slab.So, the net force on slab is$F_{net}=40$

$ma=40$

$40\times a=40$

$a=1m{s}^{-2}$

Net force on block is

$F_{net}=60N$

$ma=60$

$10\times a=60$

$a=6m{s}^{-2}$

The acceleration will be in the direction of the motion of force, thus the acceleration will be given by $\left(1.0m/s^2\right)\hat{i}$