An archer pulls back 0.75 m on a bow which has a stiffness of 200 N/m. The arrow weighs 50 g. What is the velocity of the arrow immediately after release?
The bow can be treated as a type of spring. The potential energy of a spring is: (12)kx2,
Therefore,the potential energy PE of the bow is:
PE = (12)(200)0.752 = 56.25 J
The kinetic energy of a particle is (12)mv2, The arrow can be treated as a particle since it is not rotating upon release.
Therefore, the kinetic energy KE of the arrow is:
KE = (12)(0.05)v2
If we assume energy is conserved, then
PE = KE
Solving for the velocity of the arrow v we get
v = 47.4 m/s.