Physics, 28.10.2020 17:00 alleshia2007

A 100-g bullet is fired from a rifle having a barrel 0.600 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 15 000 1 10 000x 2 25 000x 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 1.00 m long, how much work is done, and (c) how does this value compare with the work calculated in part (a)?

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Answer from: ari10184

a) W=9000J

b) W=11,666.67 J

c) The work done by the gas with a 1 meter barrel is 2,666.67 J greater than the work done by the gas with a 0.600 m barrel. This is about 1.3 times greater.

Explanation:

(The function written in the problem isn't too clearly written, but I believe it should be like this:

A 100-g bullet is fired from a rifle having a barrel 0.600 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 $15 000 + 10 000x - 25 000x^{2}$ , 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 1.00 m long, how much work is done, and (c) how does this value compare with the work calculated in part (a)?)

Since the force applied to the bullet is going to vary with the length of the barrel, we will need to make us of integration to find the work done by the gas.

So first, we start by drawing a sketch of what the problem looks like. (See attached picture). This will help us visualize the problem better.

Now, work is defined as the following integral:

$W=\int\limits^a_b {F} \, dx$

The problem tells us the force exerted by the expanding gas on the bullet is:

$F(x)=15,000+10,000x-25,000x^{2}$

So we go ahead and substitute this function into the integral.

$W=\int\limits^{0.6}_0 {15,000+10,000x-25,000x^{2}} \, dx$

Now, before integrating that function, we can factor it so it is easier for us to integrate and to deal with the values, so we can factor a common 5,000 out, so we get:

$W=5,000\int\limits^{0.6}_0 {3+2x-5x^{2}} \, dx$

and we proceed to integrate so we get:

$W=5,000[3x+x^2-\frac{5}{3}x^{3}]\limits^{0.6}_0$

and evaluate the integral, so we get:

$W=5,000([3(0.6)+(0.6)^2-\frac{5}{3}(0.6)^{3}]-[3(0)+(0)^2-\frac{5}{3}(0)^{3}]$

Which yields:

W=9000J

For part b we follow the exact same procedure but we change the limits of integration for 0 and 1 so the integral looks like this:

$W=\int\limits^1_0 {15,000+10,000x-25,000x^{2}} \, dx$

and we get the following

$W=5,000[3x+x^2-\frac{5}{3}x^{3}]\limits^1_0$

and

$W=5,000([3(1)+(1)^2-\frac{5}{3}(1)^{3}]-[3(0)+(0)^2-\frac{5}{3}(0)^{3}]$

W=11,666.67 J

c) The difference between the two works we calculated is:

11,666.67J-9000J=2,666.67J

so the work done by the gas with a 1 meter barrel is 2,666.67 J greater than the work done by the gas with a 0.600 m barrel.

if we did the following division:

$\frac{11,666.67}{9,000}=1.3$

So we can see that this is about 1.3 times greater.

A 100-g bullet is fired from a rifle having a barrel 0.600 m long. Choose the origin to be at the lo

Answer from: Quest

trig 15.3 cos

Answer from: Quest

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