An arrow is launched from a bow. The arrow accelerates over a distance of 0.415 m on the bow before leaving the bowstring at a speed of 48 m/s. The mass of the arrow is 80.0 grams. What is the average force that the bowstring exerts on the arrow. Solution Using Work-energy theorem: W = dKE Work-done is given by: W = F*d dKE = KEf - KEi = 0.5*m*Vf^2 - 0.5*m*Vi^2 Vi = 0 m/sec So, F*d = 0.5*m*Vf^2 F = 0.5*m*Vf^2/d m = 80 gm = 80*10^-3 kg Vf = final speed of arrow before leaving bowstring = 48 m/sec d = distance traveled by arrow = 0.415 m Using given values: F = 0.5*80*10^-3*48^2/0.415 F = 222.1 N Please Upvote. .