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JEE Main 2022 (24 Jun, Shift 1) · Work by a Variable Force

Work by a Variable Force question from JEE Main 2022 (24 Jun, Shift 1)

Q12JEE Main 2022 · 24 JUN · SHIFT 1Medium

Work by a Variable Force

A particle experiences a variable force F⃗=(4xi^+3y2j^)\vec{F} = \left(4x\hat{i} + 3y^2\hat{j}\right) in a horizontal x-y plane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2, 3) in the x-y plane, then Kinetic Energy changes by:

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Answer: (c)

  1. ΔK=W=∫Fx dx+∫Fy dy\Delta K = W = \int F_x\,dx + \int F_y\,dy
  2. ∫124x dx=2(4−1)=6 J\int_1^2 4x\,dx = 2(4 - 1) = 6\text{ J}
  3. ∫233y2 dy=27−8=19 J\int_2^3 3y^2\,dy = 27 - 8 = 19\text{ J}
  4. ΔK=6+19=25 J\Delta K = 6 + 19 = 25\text{ J}, so the answer is (c).
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