A block of mass $5\, kg$ is on a rough horizontal surface and is at rest. Now a force of $24\, N $is imparted to it with negligible impulse. If the coefficient of kinetic friction is $0.4$ and $g = 9.8\,m/{s^2}$, then the acceleration of the block is ........ $m/s^2$
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(d) Net force = Applied force -Friction force
$ma = 24 - \mu \;mg$
$ = 24 - 0.4 \times 5 \times 9.8$ $ = 24 - 19.6$
$⇒$ $a = \frac{{4.4}}{5} = 0.88\;m/{s^2}$
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