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JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2013
- Hard
- MCQ
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 \(g\, = 10\, ms^{-2}\))
Pick an option to check your answer.
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
\(P\; = 2Fv = 2 \times mg \times \sqrt {2g} \)
\( = 2 \times 70 \times 10 \times \sqrt {20} \)
\( = 6260.8 = 6.26 \times {10^3}\;W/s\)
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