A 60.0-kg person jumps onto the floor from a height of 3.00 m and lands stiffly, with knee joints compressing 0.500 cm. Using energy considerations and g = 9.80 m/s^2, what force acts on the knee joints?

1. Falling through height h converts gravitational potential energy mgh into kinetic energy, and the floor's force F then removes that kinetic energy over the small stopping distance d. 2. Equating the work done by the floor to the kinetic energy gained in the fall gives F d = mgh, so F = mgh / d. 3. Substituting the values, F = (60.0 kg)(9.80 m/s^2)(3.00 m) / (5.00 x 10^-3 m), converting 0.500 cm to 5.00 x 10^-3 m. 4. The numerator, mgh, is 1764 J, and dividing by the tiny stopping distance gives F = 3.53 x 10^5 N. 5. 1764 N is just the numerator mgh left undivided by the stopping distance, and 3.53 x 10^3 N and 1.76 x 10^4 N each misplace the decimal point relative to the correctly divided force. _Source: OpenStax College Physics (CC BY 4.0), Ch 7 "Work, Energy, and Energy Resources", section 7.3 Gravitational Potential Energy_