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  • A 70 kg man leaps vertically into air from a crouching position . To take the leap the man pushed the ground with a constant force F to raise himself . The centre of gravity rises by 0.5 m before h

A 70 kg man leaps vertically into air from a crouching position . To take the leap the man pushed the ground with a constant force F to raise himself . The centre of gravity rises by 0.5 m before he leaps. After the leap the c.g. rises by another 1m. The maximum power delivery by the muscle is : (Take\: \: g=10ms^{-2})

Option: 1

6.26\times 10^{3} Watts at the start .


Option: 2

6.26\times 10^{3} Watts at take off .


Option: 3

6.26\times 10^{4} Watts at the start .


Option: 4

6.26\times 10^{4} Watts at take off.


Answers (1)

best_answer

According to energy conservation , Let be take off speed v.

So,

\begin{aligned} &\frac{1}{2} m v^{2}=m g h ; \quad \text { Here } h=1 m\\ &v=\sqrt{2 g} \end{aligned}

Now Maximum power delivered by muscles  at take-off is given by

P=2 F v=2 \times m g \times \sqrt{2 g}=2 \times 70 \times 10 \times \sqrt{20}=6260.8=6.26 \times 10^{3} W / s

Hence The maximum power delivered by mussels is  6.26 * 10^{3} \mathrm{J} / \mathrm{s}

 

Posted by

Ajit Kumar Dubey

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