Question

A catapult used to hold a stone of mass 500g is extended by 20cm with an applied force F. If the stone leaves with a velocity of 40ms^-1 the value of F is

Answer 1

According to Newton's second law,

$$F=ma,\\\space\\ a=\frac{v^2}{2d},\\\space\\ F=\frac{v^2m}{2d}=2000\text{ N}.$$