physics question...seems easy but I dont know where to start...?

Question

Here's the question:
A 0.075kg arrow is fired horizontally. The bowstring exerts an average force of 65N on the arrow over a distance of .90m. With what speed does the arrow leave the bow?

I'm not sure where to start... I thought I used the
equation 1/2(m)Vfinal^2 - 1/2(m)Vintial^2 but I dont know the final velocity.

Answer 1

You start with F = ma to get acceleration

a = 65N/0.075kg = 867 m/s^2

Then you get the time from equation

d = 1/2 at^2 + vt

since initial velocity v = 0

d = 1/2at^2
t = sqrt(2d/a)

and d = 0.9m and a = 867 m/s^2

t = sqrt(2x0.9/867) = 0.046s

Then use v = at to get velocity

v = 867 x 0.046 = 40 m/s
________________
Note: I rounded everything to 2 or 3 significant digits to match the precision of your input numbers.

Answer 2

The simplest solution is to note that the kinetic energy of the arrow (mv^2 / 2) must be equal to the work done by the bowstring (0.9 x 65 Nm), and solve for v.

Answer 3

find accleration of the arrow

F = ma
65N = (.075kg)a
a = 866.67m/s^2

the arrow will acclerate until it reaches the end of .90m. After passing .90m, there is no force act on it, that means it will travel at a constant speed.

Vf^2 = 2ad + Vi^2

Vf^2 = 2(866.67)(.9) + 0^2
Vf^2 = 1560
Vf = 39.49m/s