Work Energy Theorem of a fired arrow

[Notyou]

Homework Statement

A 0.065 kg arrow is fired horizontally. The bowstring exerts an average force of 70 N on the arrow over a distance of 0.90 m. With what speed does the arrow leave the bow?

Homework Equations

W=Fd
W=KE final - KE initial
KE= 1/2mv^2

The Attempt at a Solution

First I used the force given and the distance given and found the work to be 63J. From there I set that equal to the change in kinetic energy. I believe the final kinetic energy is 0, but if that is the case I am obtaining the wrong answer. The final answer is supposed to be 44 m/s.

 

[PhanthomJay]

Homework Statement

A 0.065 kg arrow is fired horizontally. The bowstring exerts an average force of 70 N on the arrow over a distance of 0.90 m. With what speed does the arrow leave the bow?

Homework Equations

W=Fd
W=KE final - KE initial
KE= 1/2mv^2

The Attempt at a Solution

First I used the force given and the distance given and found the work to be 63J. From there I set that equal to the change in kinetic energy. I believe the final kinetic energy is 0, but if that is the case I am obtaining the wrong answer. The final answer is supposed to be 44 m/s.

It's initial KE is zero, but why do you expect its KE as it leaves the bow to be zero? It leaves it with a certain speed, and KE is associated with that speed.

 

[Notyou]

Hmm... I suppose with KE final I assumed it meant when the arrow hits the ground. So... I had it backwards? Did I just interpret the question wrong?

 

[Notyou]

Oh geeze, I just figured out what I did wrong. I wrote it down as .65 instead of .065. Thank you a lot for your help!