A 153 g bullet is fired from a rifle having a
barrel 0
.
794 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
= 12600 N,
b
= 8420 N
/
m,
c
= 10100 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.
Answer in units of J
W = ∫F•dx = a•x + (b/2) •x² - (c/3) •x³
So f = (9.8)(153) and d=.794? And then do I solve for x? How would I solve for x? ( never taken calculus)
W = ∫F•dx = a•x + (b/2) •x² - (c/3) •x³ =
=12600•0.794 +8420•0.794²/2 -10100•0.794³/3=
=10004.4+2654.1-5055.7 =7602.8 J
That answer was wrong :(
I figured it out. The answer is 10973.29 J. You just jsd2272 your x values mixed up. Thanks for the equation though. :)
To determine the work done by the gas on the bullet as it travels the length of the barrel, we need to integrate the force equation with respect to distance.
The formula for calculating work is: work = force x distance.
Given the force equation F = a + bx - cx^2, let's plug in the given values:
a = 12600 N,
b = 8420 N/m,
c = 10100 N/m^2.
To integrate the force equation, we will find the antiderivative of each term separately.
∫ (a + bx - cx^2) dx = a∫dx + b∫xdx - c∫x^2dx
The antiderivative of dx is x, so the first term becomes a∫dx = ax.
The antiderivative of xdx is x^2/2, so the second term becomes b∫xdx = b(x^2/2).
The antiderivative of x^2dx is x^3/3, so the third term becomes c∫x^2dx = c(x^3/3).
Now let's calculate the integral of the force equation:
∫ (a + bx - cx^2) dx = ax + b(x^2/2) - c(x^3/3) + C
Where C is the constant of integration.
To find the work done by the gas on the bullet as it travels the length of the barrel, we need to evaluate the above expression from x = 0 to x = 0.794 m (the length of the barrel).
So the work done (W) is:
W = ∫(0.794 m) [(ax + b(x^2/2) - c(x^3/3))] dx
Now, plug in the values of a, b, c, and integrate:
W = ∫(0.794 m) [(12600x + 4210x^2 - 3367x^3)] dx
Evaluate this definite integral: