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JEE Main 2024 (9 Apr, Shift 2) · Work by a Constant Force

Work by a Constant Force question from JEE Main 2024 (9 Apr, Shift 2)

Q7JEE Main 2024 · 9 APR · SHIFT 2Easy

Work by a Constant Force

A force (3x2+2x−5) N(3x^2 + 2x - 5)\text{ N} displaces a body from x=2 mx = 2\text{ m} to x=4 mx = 4\text{ m}. Work done by this force is ________ J.

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

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Answer: 58

  1. Variable force: W=∫24(3x2+2x−5) dx=[x3+x2−5x]24W = \int_{2}^{4} (3x^2 + 2x - 5)\,dx = \left[x^3 + x^2 - 5x\right]_2^4
  2. =(64+16−20)−(8+4−10)=60−2= (64 + 16 - 20) - (8 + 4 - 10) = 60 - 2
  3. Work done =58= 58 J.
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