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 between the block and slab is $0.60$ and coefficient of kinetic friction is $0.40$. The $10kg$ block is acted upon by a horizontal force of $100N$. The resulting acceleration of slab (in $m/s^2$ ) (Take $g=10m/s^2$ ) is____.

Answer 1

Answer: 1

$f_{s_{\max }}=0.6N_1=0.6\times 100=60N$
To start sliding between slab and block
$F-f=10a$
and $f=40a$
$\Rightarrow F=\frac{5f}{4}=75N$

as $F>75N$, acceleration of slab will be
$a=\frac{f_k}{40}=\frac{0.4\times 100}{40}=1m/s^2$