A block of mass M = 5,kg is resting on a rough horizontal surface for which coefficient of friction

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4-2.Friction

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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

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