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JEE Main 2006 (AIEEE) · Work-Energy Theorem

Work-Energy Theorem question from JEE Main 2006 (AIEEE)

Q27JEE Main 2006 · AIEEEMedium

Work-Energy Theorem

A ball of mass 0.2 kg0.2\text{ kg} is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m0.2\text{ m} while applying the force and the ball goes upto 2 m2\text{ m} height further, find the magnitude of the force. (consider g=10 m/s2g = 10\text{ m/s}^2 ).

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Answer: (d)

  1. From start (rest) to top (rest): Whand+Wgravity=ΔK=0W_{hand} + W_{gravity} = \Delta K = 0.
  2. Ball rises 0.2+2=2.2 m0.2 + 2 = 2.2\text{ m} in total: F(0.2)=mg(2.2)F(0.2) = mg(2.2)
  3. F=0.2×10×2.20.2=22 NF = \frac{0.2 \times 10 \times 2.2}{0.2} = 22\text{ N}
  4. So the answer is (d) 22 N22\text{ N}.
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