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Starting from rest, a 60.0 kg woman jumps down to the floor from a height of 0.710 m, and immediately jumps back up into the air. While she is in contact with the ground during the time interval 0 < t < 0.800 s, the force she exerts on the floor can be modeled using the function F = 9,200t − 11,500t2What is the average force exerted by the net on her?

Respuesta :

Answer:1226.7 N

Explanation:

Given

mass of woman [tex]m=60\ kg[/tex]

height [tex]h=0.710\ m[/tex]

Force is given by

[tex]F=9200t-11500t^2[/tex]

Average force imparted is given by

[tex]F_{avg}=\frac{\int_{0}^{0.8}Fdt}{\int_{0}^{0.8}dt}[/tex]

[tex]F_{avg}=\frac{\int_{0}^{0.8}(9200t-11500t^2)dt}{\int_{0}^{0.8}dt}[/tex]

[tex]F_{avg}=\dfrac{[\frac{9200t^2}{2}-\frac{11500t^3}{3}]_0^{0.8}}{[t]_0^{0.8}}[/tex]

[tex]F_{avg}=\dfrac{2944-1962.664}{0.8}[/tex]

[tex]F_{avg}=\dfrac{981.336}{0.8}[/tex]

[tex]F_{avg}=1226.67\ N[/tex]

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