A 71.4 kg man steps off a platform 3.49 m above the ground. He keeps his legs straight as he falls, but at the moment his feet touch the ground his knees begin to bend, and, treated as a particle, he moves an additional 0.59 m before coming to rest. Treating our rigid legged friend as a particle, what is the average force his feet exert on the ground while he slows down?
Assume the acceleration while he is slowing down is constant.
Answer= 4839 N. How do you do this?
2007-02-18 05:08:36 · 2 answers · asked by soulen3 2 in Science & Mathematics ➔ Physics
That answer is the given answer. I still can't solve this.
2007-02-18 05:55:18 · update #1
Calculate the speed of a falling object dropping 3.49m in an acceleration of 9.81 m/s^2 to determine the final velocity. Then calculate the acceleration required to bring that back to zero in 0.59m. When you know the acceleration, you can calculate the force required using F=ma.
2007-02-18 05:19:47 · answer #1 · answered by poorcocoboiboi 6 · 0⤊ 0⤋
An assumption is that the force applied by the bending legs is constant throughout the deceleration. The force you suggested is the answer is incorrect. I believe you typed an eight when you should have typed a one.
Force*Distance=energy
Acceleration of Gravity=9.8M/S/S
Force=Mass*Acceleration
9.8M/S/S*3.49M*71.4Kg=2442KgMM/S/S
Force=Energy/Distance
2442KgMM/S/S/.59M=4139
2007-02-18 05:35:30 · answer #2 · answered by anonimous 6 · 0⤊ 0⤋