A 60.0 kg man jumps from a height of 3.00 m. He lands on the ground and compresses his knee joints by 0.500 cm. What is the average force exerted by the ground on his knee joints?

• Mass of the man ($m$) = $60.0 \text{ kg}$
• Initial height ($h$) = $3.00 \text{ m}$
• Compression distance ($d$) = $0.500 \text{ cm}$
• Acceleration due to gravity ($g$) = $9.81 \text{ m/s}^2$

Convert the compression distance from centimeters to meters:

$$d = 0.500 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = 0.00500 \text{ m}$$

The man starts at rest at height $h$ and momentarily comes to rest after compressing his knee joints by distance $d$. Therefore, the total change in kinetic energy is zero ($\Delta KE = 0$).
According to the work-energy theorem, the net work done on the man is equal to the change in his kinetic energy.

$$W_{net} = \Delta KE = 0$$

The net work done consists of the work done by gravity ($W_g$) and the work done by the average normal force ($F_N$) from the ground (which is the force on the knee joints) during the compression.

The total vertical distance the man's center of mass falls is $h+d$.
The work done by gravity is:

$$W_g = mg(h+d)$$

The normal force $F_N$ acts upwards, while the displacement $d$ during compression is downwards. So, the work done by the normal force is negative:

$$W_N = -F_N d$$

Setting the net work to zero:

$$W_g + W_N = 0$$ $$mg(h+d) - F_N d = 0$$

$$F_N d = mg(h+d)$$ $$F_N = \frac{mg(h+d)}{d}$$

This can also be written as:

$$F_N = mg \left( \frac{h}{d} + 1 \right)$$

$$F_N = (60.0 \text{ kg})(9.81 \text{ m/s}^2) \left( \frac{3.00 \text{ m}}{0.00500 \text{ m}} + 1 \right)$$ $$F_N = (588.6 \text{ N}) (600 + 1)$$ $$F_N = (588.6 \text{ N}) (601)$$ $$F_N = 353748.6 \text{ N}$$

$$F_N \approx 3.54 \times 10^5 \text{ N}$$

The force on the man's knee joints is:

$$3.54 \times 10^5 \text{ N}$$