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JEE Main 2013 (23 Apr, Shift 1) · Power

Power question from JEE Main 2013 (23 Apr, Shift 1)

Q40JEE Main 2013 · 23 APR · SHIFT 1Difficult

Power

A 70 kg man leaps vertically into the air from a crouching position. To take the leap the man pushes the ground with a constant force F to raise himself. The center of gravity rises by 0.5 m before he leaps. After the leap the c.g. rises by another 1 m. The maximum power delivered by the muscles is: (Take g = 10 ms−2-2)

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Answer: (b)

  1. Work-energy over the whole motion: F(0.5)=mg(0.5+1)⇒F=3mg=2100 NF(0.5) = mg(0.5 + 1) \Rightarrow F = 3mg = 2100\text{ N}, net upward force F−mg=1400 NF - mg = 1400\text{ N}.
  2. Take-off speed (rises 1 m after leaving ground): v=2g×1=20=4.47 m/sv = \sqrt{2g \times 1} = \sqrt{20} = 4.47\text{ m/s}
  3. Speed is largest at take-off, so power is maximum there: P=(F−mg)v=1400×4.47≈6.26×103 WP = (F - mg)v = 1400 \times 4.47 \approx 6.26 \times 10^3\text{ W}
  4. So the answer is (b) 6.26×1036.26 \times 10^3 W at take off.
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