Question

A cable weighing 0.8 pounds per foot is attached to a small 80-pound robot, and then the robot is lowered into a 60-foot deep well to retrieve a 7 pound wrench. The robot gets out of the well (carrying the wrench) by climbing up the cable with one end of the cable still attached to the robot. How much work does the robot do in climbing to the top of the well?

Answer 1

the work i have done so far is
Work = work done by robot+wrench plus work done by chain
= (80+7)*60 + integral .8 *x , from 30 to 60

Answer 2

As the robot pulls up on the chain , the chain moves up too, but only half as far.

Answer 3

" As the robot pulls up on the chain , the chain moves up too, but only half as far."
does that mean, rope is moving as well ? rope is being lifted + robot climbing along the rope? is that a accurate reading of your question?

Answer 4

the robot wants to get out of the well with his wrench. he doesnt care about the chain, he can cut it when he has reached the top, but he needs it to pull himself upon it

Answer 5

there will be slack of 30 feet of chain
half the slack is supported by the structure (no work done), half of it was done by the robot as he climbed up

Answer 6

there will be 30 feet of slack of chain. Half the slack is supported by the structure (no work done), half of it was done by the robot as he climbed up

Answer 7

forget about the chain. there are 2 force pairs here. Robot-Rope and Robot-Gravitational field work done by the robot = -(work done by the rope on the robot)= change in the potential energy assuming the robot is still at the end of the journey.

Answer 8

no thats not the answer. try again, lol

Answer 9

here is a picture of the scenario
http://www.twiddla.com/1078807

Answer 10

@perl U posted here, you need to accept help that people have to offer. Anyway, how do u know, and if u do know, why is this question still open?

Answer 11

no i dont need to accept a half azz answer, and its wrong.

Answer 12

it wasnt even an attempt at a solution.

Answer 13

Then why did u ask? (u make no sense, u should just try to ask ur teacher or somthing)

Answer 14

i came for help. whats your problem?

Answer 15

anyways, here is the solution
http://answers.yahoo.com/question/index?qid=20120421180816AA6sTZ8

Answer 16

My problem? If u know the answer, tell us!

Answer 17

Ok, nevermind... u did tell us a sec ago. anyway, if you knew the answer before, why did u leave this question open?

Answer 18

because i dont agree with yahoo answers ,

Answer 19

it might be wrong

Answer 20

@perl I can imagine this scenario as follows:
Rope is being pulled using a pulley or something and simultaneously the robot climbs up the rope using friction(grip like action). if the rate of rope being pulled up = rate of ascent of robot with respect to the rope, then what you said is possible. i.e., by the time robot travel 30 ft along the rope, it will have reached the rope = 60 ft travel. but here is the deal, irrespective of all this the robot traveled a distance of 60 ft by pulling itself up applying a force equal to it weight or more. which would give us weight x distance. you don't have to integrate anything here. Had it been about work done by gravity on the whole or the rope, you need that.

Answer 21

Anyway, Aani probably didnt want to give u the whole answer, she probably figured u were smart enough to figure it out on ur own, was he/she wrong? And about ur "i dont agree with yahoo answers" Why did u post that here and then tell Aani he/she was wrong?

Answer 22

lily, you ask too many questions. figure it out , or post it on the side. you cant ask a question in my question

Answer 23

i dont mean to be rude, but youre out of order. this is my question

Answer 24

u r rude, but not about what u said 2 me. u dont make sense, and it is making me mad, so im out of here.

Answer 25

no, this is my question . it would be rude if it was your question

Answer 26

i doubt aani even knew the solution, so shut up

Answer 27

@Aani
here is what I think the answer is
(80+7)*60 + integral .4x on [30,60]

Answer 28

good bye.

Answer 29

aani, you do have to integrate, unless youre using some sort of center of gravity

Answer 30

the robot is climbing up the rope using his own energy (work). he is not being pulled up .

Answer 31

@perl Yes, Your answer would be correct if the robot is climbing twice as fast as the rope being pulled up. in which case we have to add the weight of tailing rope as well. But that's not that question, is it?

Answer 32

aani, this problem has NOTHING to do with the speed at which the robot is climbing. work = force * distance,

Answer 33

aani, maybe you should leave if you dont have a clue whats going on , thanks

Answer 34

goodbye lillybeth, stop typing

Answer 35

Yes, You are right. I misread the question.

Answer 36

@perl just noticed this, DO NOT tell me to shut up. :-O .... :-( jerk. u dont know how many times i wanted to tell u to shut up, but i didnt just to be a bit nicer. By the way, perl, maybe its u who doesnt have a clue about what u r talking about. Shut up and be a little kinder in ur responses. :-) (sigh... satisfaction.)

Answer 37

this is my question, you are rude telling me how to ask a question and judging me. read your comments . aani made a mistake and was a man about it. what about you>?

Answer 38

@perl @lillybeth123 No big deal people. It's simple problem.

Answer 39

why do i have to be kind, . this is my question, ok. i help people plenty, i dont have to be kind. im neutral

Answer 40

lilly, i guess you havent done many math problems, you dont get it. this is not a time or place to hold hands and such

Answer 41

you are out of line to tell me that i cant tell someone he was wrong. and HE ADMITTED IT < ROFL

Answer 42

oh shut up. Im leaving now, i will come back if i find the answer.

Answer 43

no i wont shut up, youre out of line again. this is my question

Answer 44

I dont have to stop typing, adios!

Answer 45

you said that 10 minutes ago

Answer 46

ur agravating.

Answer 47

lily, thats your opinion. i am logical

Answer 48

@perl Are you sure you've got what you wanted? I am not so sure. Do you understand it completely?

Answer 49

no, i have a different answer , no worries man

Answer 50

i just wanted to know if its right. i think its wrong

Answer 51

but you dont seem to be an expert on this, so i cant take your solution

Answer 52

oh no , lilly has returned -ducks for cover-

Answer 53

Um, guys, I posted ur question at another homework site, and i know the people pretty well, and they usually answer pretty quik.

Answer 54

shut up perl, im trying to help. u obviously dont know how to do it. and my name is not lilly.

Answer 55

lilly, what site ?

Answer 56

lilly, your handle is lillybeth. how would i know your *real* name

Answer 57

u could call me by my real name, because it is in my profile, smart alec.
go to : http://www.freemathhelp.com/forum/threads/79327-I-m-not-sure-what-this-is-really-important-please-help!?p=327786#post327786

to see ur qustion.

Answer 58

you cant expect me to know your real name, thats all im saying. right?

Answer 59

yes I can, go 2 my profile. p.s. Hannah

Answer 60

ok , one sec

Answer 61

ok hannah , thanks for posting the question. that was very nice of you .

Answer 62

:-)

Answer 63

do they respond quickly

Answer 64

this may not be the right forum though

Answer 65

usually. u should keep checking on it though. I think its close enough to the right forum. They'll answer. im pretty sure.

Answer 66

ok , thanks. i might close this question and post it again. im gonna take a break ,
do you need help with math?

Answer 67

Nope. See ya later.

Answer 68

lilly make me your fan so i can message you

Answer 69

I did.