Question

72.1 kg man steps off a platform 3.05 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 0.640 m before coming to rest.

A. What is his speed at the instant his feet touch the ground?

B. Treating him as a particle, what is the magnitude and direction of his acceleration as he slows down if the acceleration is constant?

C. Draw his free-body diagram. In terms of the forces on the diagram, what is the net force on him?

D. Calculate the average force his feet exert on the ground while he slow down. Express this force both in newtons and as a multiple of his weight.

Answer 1

A. The speed:

$$v=\sqrt{2gh}=7.7\text{ m/s}.$$

B. The acceleration acts upward, the magnitude is

$$a=\frac{v^2}{2h}=0.46\text{ m/s}^2.$$

C. The diagram:

$$F_\text{net}=F_\text{legs}-mg.$$

D. The force:

$$F_\text{legs}=F_\text{net}+mg=ma+mg=m(a+g),\\ F_\text{legs}=72.1(0.46+9.8)=740\text{ N},\\ F_\text{legs}=740/mg·mg=1.05mg.$$