Solving Work Done by Gas on a Bullet in a Rifle Barrel

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To determine the work done by gas on a bullet in a rifle barrel, the force exerted is given by F = a + bx - cx^2, where constants a, b, and c are provided. The problem requires treating the force as variable, necessitating the use of integration to calculate work done over the barrel's length. The integral W_F = ∫ F dx from 0 to the barrel length (0.796 m or 1.136 m) represents the area under the force vs. displacement graph. This approach accounts for the changing force as the bullet travels through the barrel. The discussion emphasizes understanding variable forces and the application of integration in solving the problem.
JamesL
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Heres the problem:

A 103 g bullet is fired from a rifle having a barrel .796 m long. Assuming the origin is placed where the bullet begins to move, the force exerted on the bullet by the expanding gas is F = a + bx - cx^2, where a = 10600 N, b = 5000 N/m, c = 39400 N/m^2, with x in meters.

Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Also, if the barrel is 1.136 m long, how much work is done?

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I wasnt sure whether or not to treat this problem as a variable force one... the only way i really know how to do that is by using a "spring-like" constant, but I am not sure if that is really applicable in this problem.

Could anyone point me in the right direction?
 
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Yes, the force is not constant in this problem so you have to treat it as a variable force problem. If you draw a graph of F as a function of X, what would it look like? What would the area below the graph represent? How can you find that area? Hint:
W_F = \int _{x_1}^{x_2}Fdx
 
ya...u can think of it this way...the force of the expanding gas on the bullet changes because it depends on the displacement x and the x iz changing...x changes frm what to what? 0 to .796...[so the integral can be interpreted az the force [F] changing with respect to x [dx] from x1 to x2]

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