Question

A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travels in the +x-direction. The force exerted by the expanding gas on the bullet is (16,000 + 10,000x − 26,000x2) N, where x is in meters. A) Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.
B) If the barrel is 1.05 m long, how much work is done?

Answer 1

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.

Answer 2

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;

$W_{l=0.54} = \displaystyle \int\limits^{0.54}_0 {F(x)} \, dx = \int\limits^{0.54}_0 {16,000 + 10,000 \cdot x - 26,000 \cdot x^2} \, dx$

$\displaystyle \int\limits^{0.54}_0 {16,000 + 10,000 \cdot x - 26,000 \cdot x^2} \, dx = \left[16,000\cdot x + 10,000 \cdot \frac{x^2}{2} + 26,000 \cdot \frac{x^3}{3} \right]^{0.54}_0$

Which gives;

$W_{l=0.54} = \left[16,000\cdot x + 10,000 \cdot \dfrac{x^2}{2} + 26,000 \cdot \dfrac{x^3}{3} \right]^{0.54}_0$

Therefore;

$W_{l=0.54} = \left(16,000\times 0.54 + 10,000 \times \dfrac{0.54^2}{2} + 26,000 \times \dfrac{0.54^3}{3} \right) - 0 = 11,462.688$

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;

$W_{l = 1.05} = \displaystyle \int\limits^{1.05}_0 {F(x)} \, dx = \int\limits^{1.05}_0 {16,000 + 10,000 \cdot x - 26,000 \cdot x^2} \, dx$

Which gives;

$W_{l=1.05} = \left[16,000\cdot x + 10,000 \cdot \dfrac{x^2}{2} + 26,000 \cdot \dfrac{x^3}{3} \right]^{1.05}_0$

Therefore;

$W_{l=1.05} = \left(16,000\times 1.05+ 10,000 \times \dfrac{1.05^2}{2} + 26,000 \times \dfrac{1.05^3}{3} \right) - 0 = 32,345.25$

The work done when the barrel length is 1.05 meters = 32,345.25 J

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