#1

**A 135 g bullet is fired from a rifle having a barrel 0.610 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 19000 + 9000x - 22000x^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.

(a) Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

I know what to do (integrate and then just plug in the distance) yet it says that my answer is still wrong. Is there something i'm doing wrong?

Btw this is online homework.

#2

Whats the answer supposed to be?

#3

#4

Well, let me think about this.

Do you have acceleration?

Do you have acceleration?

#5

no acceleration. It's just work problems but for some reason it says i dont have the right answer

#6

well since work is the integral of force your work function is W = 19000x + 9000x^2/ 2 - 22000x^3 / 3 so yeah then just plug and chug for your distance. ah good ol' physics mechinics. this is assuming constant acceleration of course, which indicates constant velocity since you didn't give either.

*Last edited by slash16x at Nov 25, 2008,*

#7

i did exactly that and it says that "your answer is off by a multiple of 10"

dag technology..

i got about 11599.92

dag technology..

i got about 11599.92

#8

All you ever need to know about guns!

#9

really well try again, see if you come up with something different. i'm a mechanical engineering major so i'm pretty good with physics. if it doesn't work make sure you wrote down the problem correctly. i've done that a couple times. give me a sec and i'll shoot you back my answer

#10

i know i did that right. and also i'm only allowed 3 submissions and i already used 2 X(

#11

Notice how the equation is quadratic?

That's probably not a coincidence.

That's probably not a coincidence.

#12

yeah no idea I got 11599.93 J. IDK why that's a wrong answer, I mean I did get an A in physics mechanics last semester. maybe we'll try this, put the answer in millimeters aka multiply the distance by 1000 then divide your final answer by to get the answer in meters again. let me see how that works. and btw off by a multiple of 10 up or down by 10

#13

where are you doing this homework?

#14

doesnt say just a multiple of 10

i think i'm going to stick with it except round it to 11600..if it doesnt go imma tell him how crappy this online stuff is.

EDIT: webassign .com

oh and when i multiplied by 1000 and then divided the answer by 1000 it was really small (or big).

i think i'm going to stick with it except round it to 11600..if it doesnt go imma tell him how crappy this online stuff is.

EDIT: webassign .com

oh and when i multiplied by 1000 and then divided the answer by 1000 it was really small (or big).

#15

Notice how the equation is quadratic?

That's probably not a coincidence.

yayayayaya use the quadratic exuation, find x, and then you will have your Force, and then use W=Fd

which since i am in advanced reply i cannot see

but thatll get you work.

#16

scratch that IDK why it says your answer is wrong, I got the wrong answer because work is the integral of force but maybe try multiplying 11599.92 J *.135kg the mass of the bullet see what that does. if my answer is wrong i'm stumped sorry bro.

#17

gah its ok. i tried it and got locked XD so no more submissions. whatever we had a problem last night and complained. he even said he had trouble with the system. but he said to still try the homework..

Thanks though!

Thanks though!

#18

yeah i can't think of anything else but talk to the teacher and say this problem was bad. and whoever said to solve the quadratic and use W = Fd since we have a force function that continuous that seems stupid if we can just integrate it and solve for x.

#19

Ok so you integrate between 0 and 0.61 to get the total force (11600N), but dont you then need to multiply by distance? (Ew = Fd)

so 11600*0.61 = 7076 J

Scottish Higher physics and Advanced higher mechanics, dunno how reliable that is to you yanks

so 11600*0.61 = 7076 J

Scottish Higher physics and Advanced higher mechanics, dunno how reliable that is to you yanks

#20

Yeah, you needed to integrate from 0 to the length of the barrel. Mass of the bullet has nothing to do with it if you are given the fuction of x for force.

#21

Well over here in canada they dont teach intergration, and have gotten us all used to just solving quadratically.

Thats the way i would do it.

Btw, since im not fimiliar with intergrating, how do you work that? Like what do you sub in for x?

Thats the way i would do it.

Btw, since im not fimiliar with intergrating, how do you work that? Like what do you sub in for x?

#22

Explaining integration over the internet without the use of a webcam is essentially the same as using had motions to demonstrate size to a blind woman and saying "THIS BIG!"

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So, as your calculus teacher or look up some vids online.

|___________________________________|

So, as your calculus teacher or look up some vids online.

#23

yeah that was my original theory to integrate the force from 0m to .61m but TS kept telling me the answer i got was wrong

#24

yeah that was my original theory to integrate the force from 0m to .61m but TS kept telling me the answer i got was wrong

Did you multiply by distance?

Ew = Fd

#25

Did you multiply by distance?

Ew = Fd

he integrated which is multiplying by the distance.