Question

Of a stone of mass 48g is released from a catapult whose rubber stress is 12mm assuming the first constant of the rubber 200nm-1 determine the velocity with which the stone leave the catapult

Answer 1

$$0.5mv^2=0.5kx^2\\0.048v^2=(200)(0.012)^2\\v=0.77\frac{m}{s}$$