Work question given below...

Question

xishem · Answer

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 fore (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 the barrel is 1.00m long. How much work is done.

(c) How does the value compare with the work calculated in part (a)?

xishem · Answer

I guess, given that...
$$W=\int\limits \vec F \cdot d \vec r$$And that, once again...$$dr=\hat i dx$$Then...$$\vec F \cdot d \vec r = (15,000 + 10,000x - 25,000x^2) \cdot \hat i dx$$In this case, the i-hats don't cancel out like they did in the last one. What approach do I take here?

@JamesJ

jamesj · Answer

Yes they do because the force is only in one direction, the same direction in which the bullet travels.  So the force is F [vector] = f(x) i, where f(x) = 15,000 + stuff

Then in the dot product of the i with i is equal to 1 and disappears from the calculation.

Usually for these sorts of 1-dimensional problems, we don't even bother with the vector notation.

xishem · Answer

Does this look correct?

$$W= \int\limits_{0}^{0.600}15000(dx)+10000x(dx)-25000x^2(dx)$$$$W=15000x+5000x^2-\frac{25000}{3}x^3$$Plug in 0.600 to get...$$ W = 9000J=9.00kJ$$

jamesj · Answer

Looks right.  Notationally I wouldn't write dx in brackets.  But the integral is correct.