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JEE Main 2020 (3 Sep, Shift 1) · Work-Energy Theorem

Work-Energy Theorem question from JEE Main 2020 (3 Sep, Shift 1)

Q16JEE Main 2020 · 3 SEP · SHIFT 1Medium

Work-Energy Theorem

A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 m after leaving the machine. If the part pushing the ball applies a constant force F on the ball and moves horizontally a distance of 0.2 m while launching the ball, the value of F (in N) is (g = 10 ms–2) ____.

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

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Answer: 150

  1. KE given by the machine becomes PE at the top: F×0.2=mghF \times 0.2 = mgh.
  2. mgh=0.15×10×20=30 Jmgh = 0.15 \times 10 \times 20 = 30\text{ J}
  3. F=300.2=150 NF = \frac{30}{0.2} = 150\text{ N}
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