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

Step-by-step 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* 5.6+6500* 5.6* 5.6-9333.33* 5.6* 5.6* 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

Answer 2

Step-by-step 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-(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-(28000)/(3)x^3\right ]_0^(1)`

`W_2=11000+6500-(28000)/(3)`

`W_2=8166.66 kJ`

`(W_2)/(W_1)=(8166.66)/(6599.318)=1.237 kJ`