Question

A $70kg$ 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 $0.5m$ before he leaps. After the leap the c.g. rises by another $1m$. The maximum power delivered by the muscles is : (Take $g=10m{s}^{-2}$)

• A. $6.26\times 10^3$ Watts at the start
• B. $6.26\times 10^3$ Watts at take off
• C. $6.26\times 10^4$ Watts at the start
• D. $6.26\times 10^4$ Watts at take off

Answer 1

Answer: (B)

Given: mass of man, $m=70kg$; 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.
$g\to$ acceleration due to gravity $=10m/s^2$ $\Rightarrow v=\sqrt{2gh}$
As $h=1m\Rightarrow v=\sqrt{2g}=\sqrt{2\times 10}=\sqrt{20}$ maximum power delievered by the muscles
$\begin{array}{rl}P& =2\times mg\times v=2\times 70\times 10\times \sqrt{20}\\ & \Rightarrow P=6260.99W\\ & \Rightarrow P=6.26\times 10^{3}W\text{ at take-off. }\end{array}$