A 150-g bullet is fired from a rifle having a barrel 0.580 m long. Choose the or

Question

A 150-g bullet is fired from a rifle having a barrel 0.580 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 17000 + 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 = %

Explanation / Answer

given

m = 150 g = 0.15 kg

F = 17000 + 10000 X - 26000 X^2

length of barrel or displcement for bullet is   0.58 m.

we   know that

W = Int (F.ds)

part(a)

so we have to integrate the above equation from 0 to 0.580 .

W = int (0 to 0.580 ) (17000 + 10000 X - 26000 X^2) .ds

W = (17000 X + ((10000 X^2)/2) - ((26000 X^3)/3))   from 0 to 0.580

W = (17000*0.580) + ((10000*(0.580)^2)/2) - (26000*(0.580)^3)/3)

W = 9860 + 1682 - 1690.97

W = 9851.03 J

W = 9.85 kJ

part(b)

now length of bullet or displacement of bullet is 1.05 m .

W = Int (F.ds)

so we have to integrate the above equation from 0 to 1.05

W = int (0 to 1.05 ) (17000 + 10000 X - 26000 X^2) .ds

W = (17000 X + ((10000 X^2)/2) - ((26000 X^3)/3))   from 0 to 1.05

W = (17000*1.05) + ((10000*(1.05)^2)/2) - (26000*(1.05)^3)/3)

W = 17850 + 5512.5 - 10032.75

W = 13329.75 J

W = 13.32 kJ

part (c)

percent error = (( |W1 - W2| )/W1)*100

percent error = ( ( | 9.85-13.32 | )/9.85)*100

percent error = 35.22 %

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