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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 $g\, = 10\, ms^{-2}$)
$6.26\times10^3$ Watts at the start
$6.26\times10^3$ Watts at take off
$6.26\times10^4$ Watts at the start
$6.26\times10^4$ Watts at take off
Solution
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$
