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MHT CET 2025 (23 Apr, Shift 2) · Work by a Constant Force

Work by a Constant Force question from MHT CET 2025 (23 Apr, Shift 2)

Q4MHT CET 2025 · 23 APR · SHIFT 2Easy

Work by a Constant Force

A particle is displaced from point P(3 m,4 m,5 m)P(3\text{ m}, 4\text{ m}, 5\text{ m}) to a point Q(2 m,3 m,4 m)Q(2\text{ m}, 3\text{ m}, 4\text{ m}) under a constant force F⃗=(3i^+4j^+5k^) N\vec{F} = (3\hat{i} + 4\hat{j} + 5\hat{k})\text{ N}. The work done by the force in this process is

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Answer: (d)

  1. W=F⃗⋅s⃗W = \vec{F}\cdot\vec{s} with s⃗=r⃗Q−r⃗P\vec{s} = \vec{r}_Q - \vec{r}_P.
  2. s⃗=(2−3)i^+(3−4)j^+(4−5)k^=−i^−j^−k^\vec{s} = (2-3)\hat{i} + (3-4)\hat{j} + (4-5)\hat{k} = -\hat{i} - \hat{j} - \hat{k} m
  3. W=3(−1)+4(−1)+5(−1)=−12W = 3(-1) + 4(-1) + 5(-1) = -12 J
  4. So the answer is (d) −12-12 J.
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