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JEE Main 2019 (9 Jan, Shift 2) · Work-Energy Theorem

Work-Energy Theorem question from JEE Main 2019 (9 Jan, Shift 2)

Q19JEE Main 2019 · 9 JAN · SHIFT 2Medium

Work-Energy Theorem

A force acts on a 2 kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds?

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Answer: (d)

  1. v=dxdt=6tv = \frac{dx}{dt} = 6t: v(0)=0v(0) = 0, v(5)=30 m/sv(5) = 30\text{ m/s}
  2. W=ΔK=12(2)(302)=900 JW = \Delta K = \frac{1}{2}(2)(30^2) = 900\text{ J}
  3. So the answer is (d) 900 J.
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