Question

A cable that weighs 4 lb/ft is used to lift 850 lb of coal up a mine shaft 700 ft deep. Find the work done.
(looking just for the Riemann sum and Integral setup, I'll find the answer no problem from there, and will give a medal for doing so)

Answer 1

\[\int\limits_{0}^{700}(4x) dx \]
is incorrect

Answer 2

Clearly missing something, and X\(_i\) isn't taking as... (writes)

Answer 3

cable weight is 4*700

Answer 4

so total weight is 4*700 +850

Answer 5

\[\lim_{n \rightarrow \infty} \sum_{i=1}^{n} (4)(x_i*) \triangle x\]

Answer 6

Is there something wrong with my sum there?

Answer 7

can i offer an alternative solution? if you take the average weight of the cable. i.e 1400. and then add that to the 850, you will end up with the average weight you have to lift. I would then work out the work done from there

Answer 8

should be probably 4*(700-x)

Answer 9

\[Work = \int\limits_{a}^{b} force\cdot distance\]
Correct?

Answer 10

Well and cos \(\theta\), but we don't have that for this problem

Answer 11

\[\int\limits_{0}^{700}4(700-x) + 850 dx\]

Answer 12

\[\int\limits_{0}^{700}(850-4x)dx \] I think this is your final answer

Answer 13

850 is the weight that is there till the end. The weight of cable is changing.

Answer 14

you are resting the cable weight from the coal weight...?

Answer 15

i think my setup is correct, just express lb in newtons

Answer 16

1 lb-f ≈ 4.45 N

Answer 17

so should be:
\[4,45\int\limits_{0}^{700}[4*(700-x)+850]dx\]

Answer 18

I agree @myko, however there is a much easier way to solve this

Answer 19

Got it! You were dead on mathsolver and much to my disbelief it was because I had this:
\[\lim_{n \rightarrow \infty} \sum_{i=1}^{n} (4)(x_i*) \triangle x - 850*700\]

Answer 20

It's supposed to be added >_<

Answer 21

And now it's correct, what do you know lol

Answer 22

Thank you all for helping!