highschool physics: The law of conservation of mechanical energy

Question

An archer draws back a bowstring with an average force of 500 N to a distance of 20.0 cm from equilibrium, outline the energy conversions. What is the maximum speed an 80.0 g arrow could be shot from this drawn bowstring? Answer:(50m/s) please show steps

Answer 1

Initially, prior to pulling the bowstring, the system has zero energy. Energy is input into the system in the form of work as the bow is stretched in the amount defined by W = Fd = 500(0.2) = 100 joules. This energy is stored in the bow in the form of elastic potential energy defined by E_P = 1/2kx² where "k" is the spring constant of the bow and "x" is the distance compressed (0.2 m). The amount of energy stored in the bow is equal to the energy put into the system (100 J). After the arrow is released, ignoring any losses, the energy is all converted into kinetic energy define by E_K = 1/2mv² where "m" is the mass of the arrow and "v" is it's velocity. The amount of kinetic energy is equal to the amount of potential energy the bow had and that is equal to the work put in (100 J).

100 = 1/2mv²

100 = 1/2(0.08)v²

100 = 0.04v²

100/0.04 = v²

2500 = v²

v = √2500 = 50 m/s