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NEET 2006 (AIPMT) · Work by a Constant Force

Work by a Constant Force question from NEET 2006 (AIPMT)

Q21NEET 2006 · AIPMTMedium

Work by a Constant Force

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Work done against friction is (Take g = 10 m/s2)

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Answer: (c)

Work done = increase in PE + work against friction.
$mgh = 2 \times 10 \times 10 = 200$ J
$W_f = 300 - 200 = 100$ J
So the answer is (c) 100 J.

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