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

Answer 1

Answer: (C)

$f=0.6\times 10\times 9.8N=58.8N$
Since the applied force is greater than $f$
therefore the block will be in motion
so, we should consider $f_k\cdot f_k=0.4\times 10\times 9.8N$
or $f_k=4\times 9.8N$
This would cause acceleration of $40kg$
block acceleration $=\frac{4\times 9.8}{40}=(0.98)im{s}^{-2}$
$=(1)im{s}^{-2}$
Acceleration of $10kg$ block
$\frac{58.8}{10}=(5.88)i=(6)im/s^2$