If you have an object with a mass of 2000 grams, and it has an average force of 635.029 kg pushing it for a distance of 6.096 meters before it becomes a free-flying projectile. How fast is the object traveling after the 6 meters and how do you calculate it.

active| newest | oldestF=ma

force = mass times acceleration

You have a problem here...a force should be in newtons, kilograms is mass, and doesn't inherently have a force. a kilogram of mass exerts a force due to gravity of 9.81newtons, because of the acceleration due to gravity; not because kilogram is a force.So you could mean a force of

635.029newtons...or you could mean the force 635.029 kilograms exerts due to gravity, which would be 635.029kg*9.81m/s

^{2}=6229.63449 kg-m/s^{2}, the units for newtons.I'll assume you mean 635.029 newtons for the remainder.

velocity:

V

_{f}^{2}= V_{i2}+2adFinal velocity squared = initial velocity squared + 2 (acceleration) * (distance)

unknown = acceleration, but since we have a force and a mass, we can say

F = ma

a = F/m

a = 635.029

kgm/s/2kg^{2}a = 317.5145 m/s

^{2}Assuming initial speed of zero, and rearranging the formula, we have

Vf = SquareRoot(0 + 2 * (a d))

Vf = SquareRoot(2 * (317.5145m/s

^{2}* 6.096m)Vf = SquareRoot(3871.136784m

^{2}/s^{2})Vf = 62.218 m/s

at 72psi

1413 pounds force...

or 6288 newtons.

ten times the force I had in my initial equation...

mother of god.

At sizes and pressures like that the friction and air resistance start to become a REALLY big deal

Also assumes the valve will open from zero to full bore instantly, while in reality most valve solutions (except burst disk) only open to about 25-45% by the time the projectile has left the barrel. This results in a MUCH lower average psi, especially when considering the huge volume of a 5" diameter barrel.

for example

60 pounds in newtons works

but

60 kilograms in newtons doesn't.

So, if you have 110psi at start, and 72 psi at end of barrel, you can LOOSELY say that you have an average of 91psi

If your barrel is say...2 inches diameter, 1 inch radius, the rear surface area is pi r squared,

which is pi 1 squared

which is 3.14 square inches.

91psi at 3.14 square inches is 285ish pounds of force 'average', more at the start, less at the end.

1 pound force = 4.44822162 newtons

285 pounds force = 1 267.74316 newtons

http://www.spudguns.org/calculators.html will do ALL the math for you :D

It came up with 272 pounds of force with the same info because I said the barrel took up some room in the chamber, about 10%, and the barrel pressure drop was 100 down to 73, average 86psi...the rest of the calculations are easy based on my first post

P_{1}·V_{1}= P_{2}·V_{2}If we take 1 to be the moment before firing and 2 to be the instant before the projectile breaks the seal on the end of the barrel we can calculate the necessary volume for the pressure vessel.

P

_{1}= 110 psi, P_{2}= 72 psi,V

_{2}= (pi/4)*(5in)^{2}+V_{1}= 19in^{2}*240in + V_{1}= 4712 in^{3}+V_{1}110psi·V

_{1}=72psi·(4712in^{3}+V_{1}) → 110psi·V_{1}=339264 in·lbf + 72psi·V_{1}38psi·V

_{1}= 339264 in·lbf →V_{1}=8928 in^{3}That is the size of pressure chamber you need, pressurized to 110 psi, in order to get 72 psi at the end of the barrel.

If you were to fire with these properties then drag and probably heat, along with other factors, would have a huge effect on the final velocity. In order to achieve anything close to the theoretical velocity ( 280 m/s) you are going to have to do some serious engineering.

Vf_{ID}≈6.244·dm/s^{3/2}(This gives a 4.631 cm diameter when you input frollard's velocity. If that is close then you know the assumption that the 635 really was in newton's. If it's not close then it means you need to run the equation above with the correct diameter so that your units are correct.)

The general function in terms of pressure (P, in Pa, N/m

^{2}), inside diameter (d, in cm), and mass (m, in kg) is:V_{f}≈.0125·sqrt((P·dm/s^{3})/m)http://en.wikipedia.org/wiki/Kilogram-force