How Does Barrel Length Affect Bullet Work?

[Joyci116]

Homework Statement

A 100-g bullet is fired from a rifle having a barrel 0.600 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in Newtons) exerted by the expanding gas on the bullet is 15 000 + 10 000x -25000x^2, where x is in meters. (a.)Determine the work done by the gas on the bullet as the bullet travels down the length of the barrel. (b.) What if? If the barrel is 1.00m long, how much work is done, and how does this value compare with the work calculated in part (a.)?

m=100g
x=0.600m
F=15 000N +10 000x-25 000x^2, x is meters

Homework Equations

F=kx
w=.5k\Deltax

The Attempt at a Solution

I did tried to find k, using F=kx.
k=F/x=15 000N+10 000x-(25 000x^2/x)
=(15 000N/x)+10 000-25 000x

I also tried to find work, using the w=.5k\Deltax equation.
w=.5(15 000N/x)+10 000-25 000x*.600

I am having trouble working this out problem.

 

[danielakkerma]

Hi,
Make sure you take the integral of the force for the work, in that:
<br /> \displaystyle W = \int_0^x{Fdx} <br />

 

[Uniquebum]

You are given the force as a function of x.
F(x) = ( 15000 + 10000x -25000x^2 )N

Work = Force * Distance --> W = Fd
Distance is simple, d = 0.6m

Force depends on the position of the bullet in the rifle. Thus you integrate the force function over distance 0m -> 0.6m.

That should give you about 9kN.

Then you just calculate W = 9000N * 0.6m which is aroundish 5400Nm.
The b part is simple after that. Hopefully i did it right :).

 

[Joyci116]

Thank you so much. One of my problem was that I didn't do was the integral of work.