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$\Delta$x

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$\Delta$x 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:
$\displaystyle W = \int_0^x{Fdx}$

 

[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.

 

[rwisz]

Can you provide any suggestions or tips to help me solve it more easily?

I would suggest approaching this problem by breaking it down into smaller parts and using the relevant equations step by step. First, let's determine the work done by the gas on the bullet as it travels down the 0.600 m long barrel. We can use the formula W = F*d, where W is work, F is force, and d is distance. In this case, the force is changing with distance, so we will need to use integration to calculate the work done. We can use the formula W = ∫F(x)dx, where F(x) is the force function and dx is the infinitesimal distance traveled.

To start, let's rearrange the given force function to be in terms of x: F(x) = 15 000 + 10 000x - 25 000x^2. Now, we can plug this into the work formula: W = ∫(15 000 + 10 000x - 25 000x^2)dx. Integrating this, we get W = 15 000x + 5 000x^2 - 8 333x^3 + C. We can now plug in the limits of integration, which in this case are 0 and 0.600 m, to find the work done by the gas on the bullet in the 0.600 m long barrel.

W = (15 000 * 0.600) + (5 000 * 0.600^2) - (8 333 * 0.600^3) - (15 000 * 0) - (5 000 * 0^2) + (8 333 * 0^3)

W = 11 250 + 1 800 - 2 400 - 0 - 0 + 0

W = 10 650 Joules

For part (b), we can follow the same steps and use the new distance of 1.00 m. Plugging this into the work formula, we get:

W = (15 000 * 1.00) + (5 000 * 1.00^2) - (8 333 * 1.00^3) - (15 000 * 0) - (5 000 * 0^2) + (