Consider an archery bow as behaving like a spring with a spring constant of 493 N/m. If an archer pulls the bowstring back for a distance of 42 cm, what is the elastic potential energy of the "loaded" bow?

Question

If the mass of the arrow is 39 grams, what is its speed as it leaves the bow?

Answer 1

PE=1/2 kx^2 x=.42m

1/2 mv^2=1/2 kx^2
v=x sqrt(k/m)