Question:
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 \(g=10\ \mathrm{ms^{-2}}\))
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 \(g=10\ \mathrm{ms^{-2}}\))
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Find the take-off speed from the 1 m rise, then the net force from the work-energy theorem over 0.5 m; power is largest where the speed is largest.
Updated On: Oct 31, 2026
- \(6.26\times10^{3}\) 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
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The Correct Option is B
Solution and Explanation
Take-off speed from the free rise of 1 m: \(v^2=2gh=2(10)(1)=20\), so \(v=\sqrt{20}\approx4.47\ \mathrm{m/s}\).
While pushing (distance 0.5 m), work-energy: \((F-mg)(0.5)=\tfrac12mv^2=\tfrac12(70)(20)=700\ \mathrm{J}\), so the net force \(F-mg=1400\ \mathrm{N}\) (constant).
With a constant net force the power \((F-mg)v\) grows with speed, so it is maximum at take off, where \(v\) is largest.
\(P_{max}=1400\times4.47\approx6.26\times10^{3}\ \mathrm{W}\) at take off.
Answer: (2) \(6.26\times10^{3}\) Watts at take off
While pushing (distance 0.5 m), work-energy: \((F-mg)(0.5)=\tfrac12mv^2=\tfrac12(70)(20)=700\ \mathrm{J}\), so the net force \(F-mg=1400\ \mathrm{N}\) (constant).
With a constant net force the power \((F-mg)v\) grows with speed, so it is maximum at take off, where \(v\) is largest.
\(P_{max}=1400\times4.47\approx6.26\times10^{3}\ \mathrm{W}\) at take off.
Answer: (2) \(6.26\times10^{3}\) Watts at take off
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