Question Question: A 95-g bullet is fired from a rifle having a barrel 0.636 m l

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Question:

A 95-g bullet is fired from a rifle having a barrel 0.636 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 13800 + 9500x - 23500x2, 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.
( ) kJ

(b) If the barrel is 0.97 m long, how much work is done?
( ) kJ

(c) How does this value compare with the work calculated in part (a)?
percent difference =( ) %

Explanation / Answer

Here ,

F = 13800 + 9500x - 23500 x^2

a)

work done = integration(F * dx) from 0 to 0.636 m

work done = integration(13800 + 9500x - 23500 x^2) dx from 0 to 0.636

work done = (13500 * x + 4750 x^2 - 23500 x^3/3) from 0 to 0.636

work done = 13500 * 0.636 + 4750 * 0.636^2 - 23500 * 0.636^3/3

work done = 8492.2 J

the work done by gas molecules is 8492.2 J

b)

for barrel to be 0.97 m

work done = integration(F * dx) from 0 to 0.97 m

work done = integration(13800 + 9500x - 23500 x^2) dx from 0 to 0.97 m

work done = (13500 * x + 4750 x^2 - 23500 x^3/3) from 0 to 0.97 m

work done = 13500 * 0.97 + 4750 * 0.97^2 - 23500 * 0.97^3/3

work done = 10415 J

the work done is 10415 J

c)

work calculated = (10415 - 8492.2) * 100/8492.2

work calculated = 22.64 %

the percentage of work calculated is 22.64 %

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