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A mass of 1 kg is kept on an inclined plane with 30° inclination with respect to the horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of 4 m/s. The work done by the frictional force in time 2 s is ______ J. (Take g=10 m/s²)

Asked in JEE Main 5th April 2nd Shift 2026 · Work done by friction

Answer: (1) 20

Step-by-step solution

Given: m=1 kg, incline 30°, assembly rises vertically at v=4 m/s for t=2 s.

The block is at rest relative to the incline, so static friction balances the down-slope component of weight: f=mg sin 30°=5 N, acting up the slope.

Displacement of the block is vertical: d=vt=8 m.

Angle between friction (up the 30° slope) and the vertical displacement is 60°.

W=fd cos 60°=5×8×1/2=20 J.

Why the other options are wrong

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