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  4. A 40 kg slab rests on a frictionless floor. A 10 kg 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 10 kg block is acted upon by a horizontal force of 100 N. The resulting acceleration of slab (in m / s 2 ) (Take g=10 m / s 2 ) is.

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Q. A $40\, kg$ slab rests on a frictionless floor. A $10\, kg$ 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 $10\, kg$ block is acted upon by a horizontal force of $100\, N$. The resulting acceleration of slab (in $m / s ^{2}$ ) (Take $g=10\,m / s ^{2}$ ) is____.Physics Question Image

Laws of Motion

Solution:

$f_{s_{\max }}=0.6 N_{1}=0.6 \times 100=60\, N$
To start sliding between slab and block
$F-f=10 a$
and $f=40 a$
$\Rightarrow F=\frac{5 f}{4}=75\, N$
image
as $F>75\, N$, acceleration of slab will be
$a=\frac{f_{k}}{40}=\frac{0.4 \times 100}{40}=1\, m / s ^{2}$