Calculating Arrow Speed with Doubled Force: Newton's Second Law

[Washable_Marker]

This question is driving me insane...

An arrow, starting from rest, leaves the bow with a speed of 25.0 m/s. If the average force exerted on the arrow by the bow were doubled, all else remaining the same, with what speed would the arrow leave the bow?

Of course, instinct says 50 m/s, but it's never that simple! I have no idea where to start with this, it seems like too much information is missing.

A general prod in the right direction would be appreciated- please don't tell me something like "use F=ma", because I *know* that. What I *don't* know is how to get the force, or the mass, or the acceleration.

Thanks =)

 

[arildno]

Hint:
Think of stretching the bow-string back as similar to compressing a spring.
DOUBLING the force, keeping the "rest" alike should be similar to doubling the compression distance; agreed?

Now, think of conservation of mechanical energy...

 

[Washable_Marker]

I still keep getting 50... grr.

Okay, so using

V2
------
V1

set equal to

Fa
-------
2Fa

I've tried cross-multiplying, but I still have too many unknowns to solve for anything. All I know is the change in velocity.

 

[Päällikkö]

I deleted my earlier message, as I couldn't quite get it edited (physicsforums.com appears sluggish to me right now), and the equation had a bad typo.
Anyways, what is "all else"? Time? Distance?

 

[Washable_Marker]

Yes, I assume "all else" means *everything* is the same... should I just pick a random value and keep it constant?

 

[Päällikkö]

Well, I assumed the distance of the acceleration would be same. As we have the average acceleration, I used the formulas:
x = x₀ + v₀t + ½at²
F = ma

If you managed to see the post I deleted, forget what I said in it :).

 

[Washable_Marker]

^^ what are x and xo being used to represent?

 

[Päällikkö]

x is the final disposition, x₀ is the disposition at the beginning. Are you not familiar with the equations above? What equations do you have at your disposal?

 

[arildno]

This question is too vague, since "all else" can be validly be interpreted in two distinct ways:
1. If times and distances are to be kept equal, it means that you have changed out the actual bowstring you used. Thus, not only the force changes, but also a material object pertinent to the problem.

This is what I now think they're after.
Then, use the work-energy theorem, and gain the factor of square root of two, rather than 2.

2. However, if you just double the force while having the SAME bow, you will effectively draw the string a doubled distance back, as I assumed in my first post.

 

[Washable_Marker]

I'm familiar with the first one being used as
d = v0t + ½at^2

and of course i know the second =)