A block of mass 25 kg is pulled along a horizontal surface by a force… | JEE Main 2025 (4 Apr, Shift 2)

Q6JEE Main 2025 · 4 APR · SHIFT 2Medium

Work by a Constant Force

A block of mass 25 kg is pulled along a horizontal surface by a force at an angle $45^\circ$ with the horizontal. The friction coefficient between the block and the surface is 0.25. The block travels at a uniform velocity. The work done by the applied force during a displacement of 5 m of the block is:

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Answer: (d)

Uniform velocity: $F\cos 45^\circ = \mu N$ with $N = mg - F\sin 45^\circ$ .
$\frac{F}{\sqrt{2}}(1 + \mu) = \mu mg \Rightarrow \frac{F}{\sqrt{2}} = \frac{0.25 \times 25 \times 9.8}{1.25} = 49$ N
$W = F\cos 45^\circ \times s = 49 \times 5 = 245$ J
So the answer is (d) 245 J (with $g = 9.8\text{ m/s}^2$ ).

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