Finding the delta x of a bow and arrow pullback

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Homework Help Overview

The discussion revolves around determining the delta x (displacement) of a bow's pullback in the context of energy conservation and kinetic energy. The original poster presents a scenario involving a bow and arrow, with specific values for weight and velocity, and seeks guidance on applying relevant equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of the work-energy principle, questioning how to correctly set up the equations. There is uncertainty about the mass of the arrow and the correct interpretation of kinetic energy in relation to the bow's pullback.

Discussion Status

The discussion is ongoing, with participants providing various interpretations and approaches to the problem. Some guidance has been offered regarding the relationship between the bow's draw weight and the energy stored, but there is no explicit consensus on the correct method or final answer.

Contextual Notes

Participants note potential missing information, particularly regarding the mass of the arrow, which affects the calculations for kinetic energy. There is also a discussion about the applicability of spring formulas to the mechanics of a bow, indicating a need for further clarification on assumptions made in the problem setup.

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Homework Statement


Find the delta x of the bow pullback. The weight of the pullback is 45 kg, the speed of the arrow as it shot out was 184mph( i converted that to velocity and got 83 m/s).

Homework Equations


I'm using K1+U1+W=K2+U2.

The Attempt at a Solution


I wasn't sure which formulas to plug into that equation... is it u1+work=k2? or work=k2. Then i didn't know what plugs into u1. work=FxD. please just show me where to start!
 
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So I ended up using work= kinectic energy, because it's the same. I got (444.8xDistance)=1/2(45.36)(82.6^2) then i got a very large distance? was that right?
 
Malcohmh said:
(444.8xDistance)=1/2(45.36)(82.6^2)
The arrow is 45.36kg?!
 
haruspex said:
The arrow is 45.36kg?!
No the pull on the bow! Sorry lol... Idk what I should plug in? Does that mean I have to solve for the mass in KE?
 
Malcohmh said:
No the pull on the bow! Sorry lol... Idk what I should plug in then
If you don't know the mass of the arrow then you cannot determine the KE achieved. Are you sure you have quoted the whole question, word for word?
 
haruspex said:
If you don't know the mass of the arrow then you cannot determine the KE achieved. Are you sure you have quoted the whole question, word for word?


Ya. I think I am missing some pieces.. from what I see towards the end of the video. :D
136 joules of KE= 1/2m(82.6^2). m=.04kg then i plug that into 444.8xDist.=1/2(.04)(82.06^2)! YES?! The pullback is .3m?!
 
Last edited:
Malcohmh said:
136 joules of KE= 1/2m(82.6^2). m=.04kg then i plug that into 444.8xDist.=1/2(.04)(82.06^2)! YES?! The pullback is .3m?!
Not quite. The 445 N is the peak draw weight. It will not be constant over the distance. Do you know the formula for PE of a spring?
 
It's (1/2)•444.8•Dist.=(1/2)(.04)(82.6^2)
Yes! Distance=.61meters!
 
Malcohmh said:
It's (1/2)•444.8•Dist.=(1/2)(.04)(82.6^2)
Yes! Distance=.61meters!
Are you saying that is the correct answer as given by the question setter?
There is a problem with applying the spring formula to drawing a bow. If we take the arc of the bow as behaving like a spring, the pull on the string does not. To do it more accurately, it would probably be simpler to think in terms of the movement of the ends of the bow, treating the arc as two rods jointed in the middle. When the two rods form an angle ##\pi-2\theta##, the torque at the joint is ##k\theta##. Knowing the draw length, with a bit of trig we can deduce the relationship between the draw weight (the max force applied to the string) and the max torque, and thus determine k. Then the stored energy would be ##\frac 12 k {\theta_{max}}^2##.
 
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