(Average force) x (draw back distance) = work done on bow
Set that equal to the arrow's kinetic energy in flight. Solve for V
You will not need to use the acceleration of gravity.
(M/2)*V^2 = 184.5*1.26 = 232.5 J
V = sqrt(2*232.5/0.26)= 42.3 m/s
An archer puts a 0.26 kg arrow to the bowstring. An average force of 184.5 N is exerted
to draw the string back 1.26 m.
The acceleration of gravity is 9.8 m/s
2
.
Assuming no frictional loss, with what
speed does the arrow leave the bow?
Answer in units of m/s
1 answer