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JEE Main 2024 (29 Jan, Shift 1) · Work by a Constant Force

Work by a Constant Force question from JEE Main 2024 (29 Jan, Shift 1)

Q9JEE Main 2024 · 29 JAN · SHIFT 1Easy

Work by a Constant Force

A block of mass $100\text{ kg}$ slides over a distance of $10\text{ m}$ on a horizontal surface. If the co-efficient of friction between the surfaces is 0.4, then the work done against friction $(\text{in } J$ ) is:

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Answer: (d)

Friction $f = \mu mg = 0.4 \times 100 \times 10 = 400$ N
Work against friction $W = f d = 400 \times 10 = 4000$ J
So the answer is (d) 4000 (taking $g = 10\text{ m/s}^2$ ).

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