Finding the delta x of a bow and arrow pullback

[Malcohmh]

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!

 

[Malcohmh]

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?

 

[haruspex]

(444.8xDistance)=1/2(45.36)(82.6^2)

The arrow is 45.36kg?!

 

[Malcohmh]

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?

 

[haruspex]

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?

 

[Malcohmh]

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?!

 

[haruspex]

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?

 

[Malcohmh]

It's (1/2)•444.8•Dist.=(1/2)(.04)(82.6^2)
Yes! Distance=.61meters!

 

[haruspex]

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 $\frac 12 k {\theta_{max}}^2$.