# Thread: arrow problem, flying horizontally

1. ## arrow problem, flying horizontally

A 0.062 kg arrow is fired horizontally. The bowstring exerts an average force of 65 N on the arrow over a distance of 0.75 m. With what speed does the arrow leave the bow?

need in m/s

is this just a simple equation problem, to find velocity?
or F = mass * acceleration?
or just need to find velocity?

the part that says horizontally, is confusing me.

2. Would it be faster or slower if you fired it down or up? The acceleration due to gravity is perpendicular to the bow acceleration when fired horizontally, no?

3. Thanks, i just used F= ma to find the acceleration where N = 65 and .062 kg was the mass.

then i used the kinematics equation of v^2 = v0^2 + 2*a * (xf - x0)

and plugged everything in to get the velocity after square rooting the sides.

4. Originally Posted by rcmango
A 0.062 kg arrow is fired horizontally. The bowstring exerts an average force of 65 N on the arrow over a distance of 0.75 m. With what speed does the arrow leave the bow?

need in m/s

is this just a simple equation problem, to find velocity?
or F = mass * acceleration?
or just need to find velocity?

the part that says horizontally, is confusing me.
Originally Posted by rcmango
Thanks, i just used F= ma to find the acceleration where N = 65 and .062 kg was the mass.

then i used the kinematics equation of v^2 = v0^2 + 2*a * (xf - x0)

and plugged everything in to get the velocity after square rooting the sides.
You can do it this way, though I suspect you are now learning about the Work-Energy Theorem.
$\displaystyle W = \Delta KE$

The work done is simply $\displaystyle W = Fs$ since the force and displacement are in the same direction. $\displaystyle \Delta KE = KE - KE_0 = \frac{1}{2}mv^2$ since the initial speed of the arrow is 0 m/s. Thus
$\displaystyle Fs = \frac{1}{2}mv^2$

Note: Here I am using a constant F. In simple cases we can get away with using an average force as a constant force over a time interval. (Perhaps in all cases. I'm trying to be conservative as I'm too tired to think through details right now.)

-Dan

5. not sure i'm following correctly, but when i used your formula for F = 1/2 * m * v^2

i get something around 45 m/s.

which is not correct. not sure what i've done there.

however, 39.6 m/s is correct.

6. Originally Posted by topsquark
$\displaystyle Fs = \frac{1}{2}mv^2$
Originally Posted by rcmango
not sure i'm following correctly, but when i used your formula for F = 1/2 * m * v^2

i get something around 45 m/s.

which is not correct. not sure what i've done there.

however, 39.6 m/s is correct.
You missed the "s" which is 0.75 m. (Check the units on both sides of the equation you posted. You will see that this cannot be a correct equation.)

-Dan