Answer:
A). 11462 Joules
B). 32345.25 Joules
Explanation: The workdone by a variable force undergoing a displacement is in d line of integral.
Find the attached file for the solution.
The work done by the gas on the bullet is observed as the kinetic energy
received by the bullet according to the law of energy conservation.
A) The work done by the gas on the bullet as the bullet travels the 0.54 m.
length of the barrel is 11,462.688 Joules.
B) The work done if the barrel length is 1.05 meters is 32,345.25 Joules.
Reason:
Known parameters are;
Length of the barrel, l = 0.5400 m
Location where the bullet begins to move = (0, 0)
Mass of the bullet, m = 125 g = 0.125 kg
Function for the force of the expanding gas, F(x), is presented as follows;
F(x) = 16,000 + 10,000·x - 26,000·x²
x = Measurement of length in meters
A) The work done by the gas on the bullet is given as follows;
Work done = Force × Distance = Area enclosed by the force graph
The enclosed area can be obtained by integration of the force function
over the distance x the force is applied as follows;
[tex]W_{l=0.54} = \displaystyle \int\limits^{0.54}_0 {F(x)} \, dx = \int\limits^{0.54}_0 {16,000 + 10,000 \cdot x - 26,000 \cdot x^2} \, dx[/tex]
[tex]\displaystyle \int\limits^{0.54}_0 {16,000 + 10,000 \cdot x - 26,000 \cdot x^2} \, dx = \left[16,000\cdot x + 10,000 \cdot \frac{x^2}{2} + 26,000 \cdot \frac{x^3}{3} \right]^{0.54}_0[/tex]
Which gives;
[tex]W_{l=0.54} = \left[16,000\cdot x + 10,000 \cdot \dfrac{x^2}{2} + 26,000 \cdot \dfrac{x^3}{3} \right]^{0.54}_0[/tex]
Therefore;
[tex]W_{l=0.54} = \left(16,000\times 0.54 + 10,000 \times \dfrac{0.54^2}{2} + 26,000 \times \dfrac{0.54^3}{3} \right) - 0 = 11,462.688[/tex]
The work done by the gas on the bullet as the bullet travels the 0.54 m. length of the barrel, W = 11,462.688 J
B) If the barrel is 1.05 m long, we have;
[tex]W_{l = 1.05} = \displaystyle \int\limits^{1.05}_0 {F(x)} \, dx = \int\limits^{1.05}_0 {16,000 + 10,000 \cdot x - 26,000 \cdot x^2} \, dx[/tex]
Which gives;
[tex]W_{l=1.05} = \left[16,000\cdot x + 10,000 \cdot \dfrac{x^2}{2} + 26,000 \cdot \dfrac{x^3}{3} \right]^{1.05}_0[/tex]
Therefore;
[tex]W_{l=1.05} = \left(16,000\times 1.05+ 10,000 \times \dfrac{1.05^2}{2} + 26,000 \times \dfrac{1.05^3}{3} \right) - 0 = 32,345.25[/tex]
The work done when the barrel length is 1.05 meters = 32,345.25 J
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