A 60.0-kg person jumps onto the floor from a height of 3.00 m. If he lands stiffly (with his knee joints compressing by 0.500 cm), calculate the force on the knee joints.

Step 1: Set up an expression for the force applied on knee joints
Given that
Mass of the person m = 60 kg
Height h = 3.00 m
Acceleration due to gravity g = 9.8 m/s^2
Knee joints compressed (displacement), d = 0.5 cm = 0.005 m
Since the force applied on the knee is upward and the direction of displacement is downward, the angle between force(F) and displacement(d) would be 180\degree.
As he comes to a halt, the work performed on the individual by the floor W = Fd \cos \theta = Fd \cos 180\degree
W = - Fd
The kinetic energy is equivalent to the work performed by the floor on the knee of a person.
W = KE = - Fd , here negative sign indicates opposite direction of the energy(KE)
The amount of potential energy lost by dropping from height is equal to the kinetic energy the individual has when they hit the ground.
KE = PE
Fd = mgh
F = \frac{mgh}{d}
Step 2: Finding the force on the knee joints
F = \frac{60*9.8*3}{0.500}
F = 352800
Hence, Applied force on the knee F = 352800 N
I found an answer from ntrs.nasa.gov
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