An arrow with a mass of 0.12 kg is pulled back in a bow. After being held still, it is let go, moving at 23m/s after the 0.51meters in which the arrow is in contact with the bow string? What work was done on the arrow by th bow string? What force was exerted by the archer to pull the bowstring back?

Answer 1

In this problem, it is best to assume that the force applied to the arrow is proportional to the distance that the arrow is pulled back. Thus an uneven force is applied, starting at Fmax and ending at zero when the string is straight up and down.

The kinetic energy given to the arrow equals the work done by the bowstring in pushing the arrow forward. This equals (1/2) Fmax * X, where X = 0.51 m

Fmax = (1/2) M V^2/(X/2) = M V^2/X

Solve fr Fmax

Answer 2

Hello, why is there a 1/2 in your kinetic formula if you are using Fmax and delta x? W= (Fmax)(delta x) and Ek =1/2 m(v^2)