Laws of Motion

JEE Main level, single correct, medium

127 of 210

Question 127 of 210

A mass of 1 kg1\text{ kg} is kept on an inclined plane with 30°30° inclination with respect to horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of 4 m/s4\text{ m/s}. The work done by the frictional force in time 2 s2\text{ s} is _______ J. (Take g=10 m/s2g = 10\text{ m/s}^2)

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Answer: (A) 2020

Since the assembly moves with a constant velocity, the acceleration of the mass is zero. The mass is in equilibrium relative to the incline.

The frictional force ff acting on the mass balances the component of gravity along the inclined plane. Therefore, the magnitude of the frictional force is:

f=mgsin⁡30∘f = mg \sin 30^\circ

Substituting the given values (m=1 kgm = 1\text{ kg}, g=10 m/s2g = 10\text{ m/s}^2):

f=1×10×12=5 Nf = 1 \times 10 \times \dfrac{1}{2} = 5\text{ N}

The direction of the frictional force is upwards along the incline. Since the incline makes an angle of 30∘30^\circ with the horizontal, the frictional force makes an angle of 90∘−30∘=60∘90^\circ - 30^\circ = 60^\circ with the vertical upward direction.

The displacement ss of the assembly in time t=2 st = 2\text{ s} is vertically upwards:

s=vt=4×2=8 ms = vt = 4 \times 2 = 8\text{ m}

The work done by the frictional force is:

W=fscos⁡60∘W = f s \cos 60^\circ

W=5×8×12=20 JW = 5 \times 8 \times \dfrac{1}{2} = 20\text{ J}

Answer: 2020

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