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JEE Main 2023 (12 Apr, Shift 1) · Work by a Constant Force

Work by a Constant Force question from JEE Main 2023 (12 Apr, Shift 1)

Q13JEE Main 2023 · 12 APR · SHIFT 1Medium

Work by a Constant Force

To maintain a speed of 80 km/h by a bus of mass 500 kg on a plane rough road for 4 km distance, the work done by the engine of the bus will be ____________ kJ. [The coefficient of friction between tyre of bus and road is 0.04.]

Numerical answer type. Work it out, then open the answer.

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Answer: 784

Constant speed, so engine force = friction $= \mu mg = 0.04 \times 500 \times 9.8 = 196$ N
$W = F d = 196 \times 4000 = 784000$ J
Work done by the engine $= 784$ kJ.

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