A student of mass M is 1.5 ,m tall and has he | vedclass

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(B) Let the initial position of the center of mass be $h_i = 1.0 \,m$.When she bends her knees,the center of mass is lowered by $0.2 \,m$,so the new p

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$2$

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(B) Let the initial position of the center of mass be $h_i = 1.0 \,m$.When she bends her knees,the center of mass is lowered by $0.2 \,m$,so the new position is $h_{new} = 1.0 - 0.2 = 0.8 \,m$.She pushes the ground with a constant force $F$ over a distance of $d = 0.2 \,m$.At the maximum height,the feet are $0.3 \,m$ above the ground. Since the center of mass is $1.0 \,m$ above the feet when standing straight,the final height of the center of mass is $h_f = 0.3 + 1.0 = 1.3 \,m$.Applying the work-energy theorem from the position where she starts pushing to the maximum height:Work done by force $F$ + Work done by gravity = Change in kinetic energy.$F 	imes 0.2 - Mg(h_f - h_{new}) = 0 - 0$.$F 	imes 0.2 - Mg(1.3 - 0.8) = 0$.$F 	imes 0.2 = Mg 	imes 0.5$.$\frac{F}{Mg} = \frac{0.5}{0.2} = 2.5$.

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