part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
[tex]W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx = [/tex]
[tex]=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3} [/tex]
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
[tex]W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} = [/tex]
[tex]=8733 J=8.73 kJ [/tex]
part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
[tex]W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =[/tex]
[tex]=12280 J=12.28 kJ[/tex]
