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4-2.Friction
hard
A block of mass $M = 5\,kg$ is resting on a rough horizontal surface for which the coefficient of friction is $0.2$. When a force $F = 40\,\,N$ is applied, the acceleration of the block will be ........ $m/\sec^2$ $(g = 10\,\,m/{\sec^2})$
A
$5.73$
B
$8.0$
C
$3.17$
D
$10.0$
Solution
(a) Kinetic friction =${\mu _k}R$$ = 0.2(mg – F\sin 30^\circ )$
$ = 0.2\left( {5 \times 10 – 40 \times \frac{1}{2}} \right)$$ = 0.2(50 – 20) = 6\;N$
Acceleration of the block $ = \frac{{F\cos 30^\circ – {\rm{Kinetic\, friction}}}}{{{\rm{Mass}}}}$
$ = \frac{{40 \times \frac{{\sqrt 3 }}{2} – 6}}{5} = 5.73\;m/{s^2}$
Standard 11
Physics
