# Homework Help: Question on Newton's Second Law of Motion

1. Sep 18, 2008

### Nivlac2425

1. The problem statement, all variables and given/known data

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

2. Relevant equations

F = ma

3. The attempt at a solution

I know I need to find a new initial velocity for the arrow for which the force is doubled. And the question states that all else remains constant. I may need to find the acceleration using F = ma but I'm just not sure where to start at all. I just need a small kick so I know where I'm going

2. Sep 18, 2008

### Dschumanji

If a force F gives the arrow an acceleration of a, what would a force of 2F accelerate the arrow by? Use Newton's second law, it's simple algebra.

3. Sep 19, 2008

### Nivlac2425

It's double the initial accceleration right? But how would I find the speed of that?

4. Sep 19, 2008

### Dschumanji

Yup.

You can assume the force is applied for the same amount of time in each case, so in each case the acceleration lasts that long. Set up some equations that relate the velocity of the arrow after t seconds for the first and second case.

5. Oct 5, 2008

### kealh

Umm i dont get what you are telling him? Would you mind explain? please.

6. Oct 5, 2008

### Dschumanji

I really think that I made an error on this problem. If the acceleration is applied for the same duration on the arrow in each case then the velocity in the second case would just be double the velocity in the first case. If the acceleration is applied over the same distance then the velocity of the of the arrow in the second case is $$\sqrt{2}$$ times more than the velocity in the first case. I don't know which one is right.

The question is very vague when it says "All else remaining the same." In reality, to get double the average force out of the bow, the distance over which the average force is applied (equal to how far back you pull the string) is not the same in both cases; you could assume that this distance can't be changed. You could keep the distance constant in both cases but that would require more tension in the bow string, which you could also assume can't be changed too. It follows that the amount of time the average force is applied would be different in each case.

Anyone else have some insight?

7. Oct 5, 2008

### Dschumanji

Really? Nobody has any input?

8. Oct 6, 2008

### kealh

Eh yea i'm still confuse. but thank though

9. Oct 6, 2008

### nasu

The confusion comes from the word "average" in the problem. You can have time average or distance average. The result depends on which one you use.

I suppose the problem may belong to a textbook chapter about momentum where you have the equation
Fave= delta (mv).
In this case the average is over time.
And is the only equation in this chapter involving an average so there is no confusion...