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haxxorboi
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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...
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)
PE=KE
Force*Distance=.5*mass*velocity^2
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
Homework Statement
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)
Homework Equations
PE=KE
Force*Distance=.5*mass*velocity^2
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