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A 50-g bullet is fired from a rifle having a barrel 0.660 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 18000 + 10000x - 26000x2, 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. (Enter your answer to at least two decimal places.) kJ (b) If the barrel is 1.05 m long, how much work is done? (Enter your answer to at least two decimal places.) kJ (c) How does this value compare with the work calculated in part (a)? percent error = %

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krlosdark12

Answer:

a) 11.57 kJ

b) 14.40 kJ

c) 24.46%

Explanation:

The work is defined as:

[tex]W=\int\limits^x_0 {F(x)} \, dx[/tex]

[tex]W=\int\limits^x_0 {(18000+10000x-26000x^2)} \, dx\\W=18000x+5000x^2-\frac{26000x^3}{3}[/tex]

for a)

[tex]W=18000*0.660+5000*0.660^2-\frac{26000(0.660)^3}{3}\\W=11.57kJ[/tex]

for b)

[tex]W=18000*1.05+5000*1.05^2-\frac{26000(1.05)^3}{3}\\W=14.40kJ[/tex]

c)

the percent error is given by:

[tex]Error_{\%}=\frac{|W_b-W_a|}{W_a}*100\\\\Error_{\%}=\frac{|14.40kJ-11.57kJ|}{11.57kJ}*100\\\\Error_{\%}=24.46\%[/tex]

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