Question

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?

Answer 1

Answer:1226.7 N

Step-by-step explanation:

Given

mass of woman
`m=60\ kg`

height
`h=0.710\ m`

Force is given by

`F=9200t-11500t^2`

Average force imparted is given by

`F_(avg)=(\int_(0)^(0.8)Fdt)/(\int_(0)^(0.8)dt)`

`F_(avg)=(\int_(0)^(0.8)(9200t-11500t^2)dt)/(\int_(0)^(0.8)dt)`

`F_(avg)=([(9200t^2)/(2)-(11500t^3)/(3)]_0^(0.8))/([t]_0^(0.8))`

`F_(avg)=(2944-1962.664)/(0.8)`

`F_(avg)=(981.336)/(0.8)`

`F_(avg)=1226.67\ N`