IIT-JEE 2010 · Paper 1 · Physics Q63

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PhysicsSingle correct+3 / -1Question 63 of 84

A block of mass m is on an inclined plane of angle $\theta$ . The coefficient of friction between the block and the plane is $\mu$ and $\tan\theta>\mu$ . The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from $P_1=mg(\sin\theta-\mu\cos\theta)$ to $P_2=mg(\sin\theta+\mu\cos\theta)$ , the frictional force f versus P graph will look like

A block on an incline of angle θ pushed up the plane by force P.

f against P: a straight line falling from a positive value at P₁ through zero to a negative value at P₂.

f against P: constant positive between P₁ and P₂.

f against P: a straight line rising from a negative value at P₁ through zero to a positive value at P₂.

f against P: constant negative between P₁ and P₂.

Answer and worked solution

Official answer: A

Equilibrium along the incline (up the plane positive):
$P+f-mg\sin\theta=0\implies f=mg\sin\theta-P.$

Range. At $P_1$ , $f=\mu mg\cos\theta>0$ (friction up the slope); at $P_2$ , $f=-\mu mg\cos\theta$ (down the slope). In between, $f$ falls linearly through zero.

(A)

Topics:Friction

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IIT-JEE 2010 · Paper 1, Physics question 63