I should preface this, this may be the stupidest question yet on this forum. But I can't get it to work out right so I'm going wrong somewhere obvious I'm sure...(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

A crossbow is readied for release. Suppose it takes 45.0 pounds of force to draw the arrow back by 13.0 inches, and the weight of the arrow is 2 ounces. What is the speed of the arrow when it is released?

(1 lb = 4.448 N; 1 in = 2.54 cm; 1 oz = 28.35 grams)

2. Relevant equations

PE=KE

Force*Distance=.5*mass*velocity^2

3. The attempt at a solution

45lb=200.16N

13in=.3302M

2oz=.0567kg

66.092832NM=66.09832J

66.09832J=.5*.0567*v^2

66.09832J=.02835*v^2

2331.5104=v^2

v=48.2857

Any help is appreciated.

Thanks

**Physics Forums - The Fusion of Science and Community**

# Crossbow bolt velocity

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Crossbow bolt velocity

Loading...

**Physics Forums - The Fusion of Science and Community**