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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

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 \(g\, = 10\, ms^{-2}\))

  1. A \(6.26\times10^3\) Watts at the start
  2. B \(6.26\times10^3\) Watts at take off
  3. C \(6.26\times10^4\) Watts at the start
  4. D \(6.26\times10^4\) Watts at take off
Verified Solution

Answer & Solution

Correct Answer

(B) \(6.26\times10^3\) Watts at take off

Step-by-step Solution

Detailed explanation

According to energy conservation , Let be take off speed \(v.\) So, \(\frac{1}{2}m{v^2} = mgh;\;\) \(Here\;h = 1\;m\) \(v = \sqrt {2g} \) Now Maxium power delivered by musheles is given by \(P\; = 2Fv = 2 \times mg \times \sqrt {2g} \)…
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