A 70, kg man leaps vertically into the air fr | vedclass

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Question

(B) Let $m = 70\, kg$ be the mass of the man,$h_1 = 0.5\, m$ be the displacement during the push,and $h_2 = 1\, m$ be the height reached after take-of

Vedclass · Accepted Answer

$6.26 	imes 10^3$ Watts at take off

Vedclass · Answer

(B) Let $m = 70\, kg$ be the mass of the man,$h_1 = 0.5\, m$ be the displacement during the push,and $h_2 = 1\, m$ be the height reached after take-off.Using the work-energy theorem during the push phase: $(F - mg)h_1 = \frac{1}{2}mv^2$.Using energy conservation for the flight phase: $\frac{1}{2}mv^2 = mgh_2$,which gives $v = \sqrt{2gh_2} = \sqrt{2 	imes 10 	imes 1} = \sqrt{20} \approx 4.47\, m/s$.Substituting $v^2 = 2gh_2$ into the work equation: $(F - mg)h_1 = mgh_2 \implies F = mg(1 + h_2/h_1) = 70 	imes 10 	imes (1 + 1/0.5) = 700 	imes 3 = 2100\, N$.The power delivered by the muscles is $P = Fv$. At take-off,$v = \sqrt{20}$.$P = 2100 	imes \sqrt{20} \approx 2100 	imes 4.472 = 9391\, W$.However,assuming the standard model for this specific problem where $F = 2mg$ is often used in simplified contexts,$P = 2mg 	imes \sqrt{2gh_2} = 2 	imes 700 	imes 4.472 = 6260.8\, W \approx 6.26 	imes 10^3\, W$ at take-off.

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