Calculating Crossbow Bolt Velocity

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Homework Help Overview

The discussion revolves around calculating the velocity of a crossbow bolt based on the force applied to draw the bowstring and the mass of the arrow. Participants are examining the relationship between potential energy and kinetic energy in this context.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the initial calculations involving force, distance, and mass, questioning the accuracy of energy conservation assumptions. There is a focus on the energy transfer during the release of the arrow and the implications of using average force versus maximum force.

Discussion Status

The conversation is ongoing, with participants providing insights into the significance of significant figures and the physical realism of the calculated results. Some express confusion regarding the feedback from the homework system and the validity of the answers provided.

Contextual Notes

There are mentions of homework system feedback and the importance of significant figures in reporting answers. Participants are also reflecting on the nature of energy conservation in the context of the problem.

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...

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
 
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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.
 
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.
 
haxxorboi said:
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.
 
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.
 
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 said:
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.
 
haxxorboi said:
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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