Questions

A block of mass 2 kg rests on a rough inclined plane making an angle of 30^{0} with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is

a

9.8N

b

10N

c

0.98N

d

98N

detailed solution

Correct option is A

$\begin{array}{l}{\mathrm{F}}_{\mathrm{l}}=\mathrm{\mu mgcos}\mathrm{\theta}\\ {\mathrm{F}}_{\mathrm{l}}=0.7\times 2\times 10\times \mathrm{cos}{30}^{\circ}=12\mathrm{N}\end{array}$

(approximately) but when the block is lying on the inclined plane then component of weight down the plane = $\mathrm{mgsin\theta}$

$=2\times 9.8\times \mathrm{sin}{30}^{\circ}=9.8\mathrm{N}$, It means the body is stationary so static friction will work on it, Static friction = applied force = 9.8N

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