JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Work, Power & Energy

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of

6 0^{\circ}

by a force of 10 N parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m along inclined surface, the work done against frictional force is:

[g = 10 m / s^{2}]

(A) $53$ J
(B)5 J✓ Correct
(C) $5 \times 1 0^{3}$ J
(D)10 J

Answer: (B)

Solution

The block moves along the incline, and friction acts opposite to the motion, so work done against friction equals

μ N \times d

.

The normal reaction balances the component of weight perpendicular to the incline:

N = m g cos 6 0^{\circ} = 1 \times 10 \times 0.5 = 5 N

With

μ_{s} = 0.1

(given in the figure) and displacement

d = 10 m

along the incline:

W_{f r i c t i o n} = μ N d = 0.1 \times 5 \times 10 = 5 J

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Work, Power & Energy →

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Work, Power & Energy

Solution

More JEE Main questions on Work, Power & Energy