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JEE Main 2023 (1 Feb, Shift 2) · Work by a Variable Force

Work by a Variable Force question from JEE Main 2023 (1 Feb, Shift 2)

Q6JEE Main 2023 · 1 FEB · SHIFT 2Easy

Work by a Variable Force

A force F=(5+3y2)F = \left(5 + 3y^2\right) acts on a particle in the yy-direction, where FF is in newton and yy is in meter. The work done by the force during a displacement from y=2 my = 2\text{ m} to y=5 my = 5\text{ m} is ___________ J.

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

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Answer: 132

  1. W=∫25(5+3y2) dy=[5y+y3]25W = \int_2^5 (5 + 3y^2)\,dy = \left[5y + y^3\right]_2^5
  2. W=5(3)+(125−8)=15+117W = 5(3) + (125 - 8) = 15 + 117
  3. Work done =132= 132 J.
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