# arrow problem, flying horizontally

• Nov 2nd 2007, 09:27 AM
rcmango
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.
• Nov 2nd 2007, 01:12 PM
TKHunny
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?
• Nov 2nd 2007, 01:58 PM
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.
• Nov 2nd 2007, 07:57 PM
topsquark
Quote:

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.

Quote:

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.) (Sleepy)

-Dan
• Nov 3rd 2007, 06:25 PM
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.
• Nov 4th 2007, 02:28 AM
topsquark
Quote:

Originally Posted by topsquark
$\displaystyle Fs = \frac{1}{2}mv^2$

Quote:

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