Question

You release a 1.4-kg ball from rest at a height of 2 m above soft ground. Once it has come to rest, the ball has made an 8-cm deep hole in the ground.
What is the magnitude of the force that the ground exerts on the ball while it is stopping?

Answer 1

The change of energy is equal to work done, so

$$\Delta E=W$$

$$mgh=Fd$$

Therefore, the magnitude of the force

$$F=\frac{mgh}{d}$$

$$=\frac{1.4\:\rm kg\times 9.8\: m/s^2\times 2\: m}{0.08\:\rm m}=343\:\rm N$$