1. Tardigrade
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  3. Physics
  4. 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)

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Q. 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 $g = 10 \,ms^{-2}$)

JEE MainJEE Main 2013Work, Energy and Power

Solution:

Given: mass of man, $m=70 kg$; Man pushes the ground with force $f$ to take the leap and his center of gravity rises by In after the leap.
Law of conservation of energy is applied. Kinetic energy initially $=$ Potential energy $\Rightarrow \frac{1}{2} m v^2=m \cdot g \cdot h$ raised after leap.
$g \rightarrow$ acceleration due to gravity $=10 m / s ^2$ $\Rightarrow v=\sqrt{2 g h}$
As $h=1 m \Rightarrow v=\sqrt{2 g}=\sqrt{2 \times 10}=\sqrt{20}$ maximum power delievered by the muscles
$ \begin{aligned} P & =2 \times m g \times v=2 \times 70 \times 10 \times \sqrt{20} \\ & \Rightarrow P=6260.99 W \\ & \Rightarrow P=6.26 \times 10^3 W \text { at take-off. } \end{aligned} $