A block is kept on a frictionless inclined surface with angle of inclination $\alpha$. The incline is given an acceleration a to keep the block stationary. Then a is equal to

(B) $\mathrm{g} \operatorname{cosec} \alpha$

JEE Main · 2005 · 08 May 2005 · Offline · Laws of Motion

A block is kept on a frictionless inclined surface with angle of inclination

α

. The incline is given an acceleration a to keep the block stationary. Then a is equal to

(A)g/tan $α$
(B) $g cosec α$
(C)g
(D) $g tan α$ ✓ Correct

Answer: (D)

Solution

Work in the frame of the accelerating incline, where the block feels a backward pseudo-force

ma

horizontally in addition to gravity. For it to stay put on the smooth surface, the component of gravity down the slope,

m g sin α

, must be balanced by the component of the pseudo-force up the slope,

ma cos α

. Hence

m g sin α = ma cos α

, giving

a = g tan α

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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