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Crossbow bolt velocity

by haxxorboi
Tags: bolt, crossbow, velocity
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haxxorboi
#1
Sep17-08, 12:56 PM
P: 24
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...

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
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mgb_phys
#2
Sep17-08, 01:11 PM
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The force is only 45lbs at the end of the pull but it is zero at the point where the arrow leaves the string.
If you draw a force/distance diagram it's easy to see that you only get half the energy you calcualted.
haxxorboi
#3
Sep17-08, 01:15 PM
P: 24
Ah geez, me and my linear thinking. Yep, that was it.

Final answer: 34.141740036184633488266425671601 m/s

Thanks for the help mgb_phys! I appreciate it.

mgb_phys
#4
Sep17-08, 02:32 PM
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Crossbow bolt velocity

Quote Quote by haxxorboi View Post
Final answer: 34.141740036184633488266425671601 m/s
No it isn't
Knowing why that is a ridiculous answer is as important as knowing how to get it.
haxxorboi
#5
Sep17-08, 02:43 PM
P: 24
I guess I don't understand what you mean. The homework system I'm using gives feedback after you complete a problem and tells you whether you're correct or not. The answer I posted here was MScalc's output that I just copy & pasted. I inputted 34.142 m/s and it returned "You are correct. The computer's answer is 34.14 m/s"

What am I missing here? I understand there will never be 100% conservation of energy to give to the bolt, however this is a pretty basic physics course and most of the questions don't deal with any outside energy losses.
mgb_phys
#6
Sep17-08, 03:04 PM
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34.141740036184633488266425671601
You know the force and distance to 3 significant figures and the mass to 1, where do the 32 significant figures in the answer come from?

As an example, the size of an atom is about 0.0000000001m, you are quoting distance to an accuracy of 0.0000000000001 of a proton!
This isn't just being picky - it is completely unphysical to quote an answer like that.
haxxorboi
#7
Sep17-08, 03:09 PM
P: 24
Quote Quote by haxxorboi View Post
The answer I posted here was MScalc's output that I just copy & pasted. I inputted 34.142 m/s and it returned "You are correct. The computer's answer is 34.14 m/s"
I was just copy and pasting the rough answer I got from the calculator. This wasn't the actual answer I used. I really don't remember too much of Sig. Digits as the only time I used them was about 3 years ago in high school chemistry but either way my professor doesn't use them.

Was my extremely lengthy answer "correct", no. But was it, in it's rounded form that is, the answer I needed, yes.

In the future I'll attempt to answer in a more realistic format. Sorry about that.
mgb_phys
#8
Sep17-08, 03:51 PM
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Quote Quote by haxxorboi View Post
In the future I'll attempt to answer in a more realistic format. Sorry about that.
No problem - it wasn't anything personal I was just trying to make a point!
It's like getting the units right, there is an attitude of "thats what the calcualtor said - it must be right" that you have to get beyond.


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