Question

A 130 g bullet is fired from a rifle having a barrel 0.560 m long. Assuming the origin is placed where the bullet begins to move, the force (in newtons) exerted by the expanding gas on the bullet is 11000 + 13000 x - 28000 x 2, 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) what if? if the barrel is 1.00m long , how much work is done, and how does this value compare with thework calculated in part a)?

Answer 1

Answer:

Explanation:

Given

mass of bullet $m=130 gm$

Length of barrel $L=0.56 m$

Force $F=11000+13000x-28000x^2$

Work done $W=\int_{x_0}^{x_f}Fdx$

$W_1=\int_{0}^{0.56}\left ( 11000+13000x-28000x^2\right )dx$

$W_1=\left [ 11000x+6500x^2-\frac{28000}{3}x^3\right ]_0^{0.56}$

$W_1=6160+2038.4-1639.082=6599.318 J\approx 6.599 kJ$

(b)  When barrel is 1 m long

$W_2=\int_{0}^{1}\left ( 11000+13000x-28000x^2\right )dx$

$W_2=\left [ 11000x+6500x^2-\frac{28000}{3}x^3\right ]_0^{1}$

$W_2=11000+6500-\frac{28000}{3}$

$W_2=8166.66 kJ$

$\frac{W_2}{W_1}=\frac{8166.66}{6599.318}=1.237 kJ$

Answer 2

Answer:

Explanation:

mass of bullet, m = 130 g

Force, F = 11000 + 13000 x - 28000 x²

(a) the work done by the gas is given by

$W = \int F dx$

$W_{1} = \int_{0}^{0.56}\left ( 11000+13000x-28000x^{2} \right )dx$

$W_{1}=11000\times 5.6+6500\times 5.6\times 5.6-9333.33\times 5.6\times 5.6\times 5.6$

W1 = 1904522.08 J

(b) the work done by the gas is given by

$W = \int F dx$

$W_{2} = \int_{0}^{1}\left ( 11000+13000x-28000x^{2} \right )dx$

[tex]W_{1}=11000\times 1+6500\times 1-9333.33\times 1

W2 = 8166.67 J

W1 / W2 = 233.2