Question

A cable that weighs 8 lb/ft is used to lift 650 lb of coal up a mine shaft 600 ft deep. Find the work done. Show how to approximate the required work by a Riemann sum. (Let x be the distance in feet below the top of the shaft. Enter xi* as xi.)

Answer 1

Answer:

1830000 ft*lb

Step-by-step explaination:

Work done = fd

f=force:

d = distance

f = 650 + 8x lbs, x is current amount of cable in shaft

dW = fdx = (650+8x)dx

x ranges from 0 to 600 then the total work is:

W = integral(x=0 to x=600)((650+8x)dx)

W = (650x+4x^2)[x=0, x=600]

W = (650*600 + 4*(600)^2) = 1830000 ft*lb

W = 1830000 ft*lb

Riemann sum approximation:

W = sum(i=1 to i=n)(650+8xi*)(xi - x(i-1))

x0 is 0, xn is 600 (or x0 is 600, xn is 0)

assuming the partition is a constant interval deltax sum can be written as:

W = summation(i=1 to i=n)(650+8xi*)deltax

Answer 2

The work done is 1830000 ft.lb

Given that a cable that weighs 8 lb/ft is used to lift 650 lb of coal up a mine shaft 600 ft deep.

We know Work done (W) = fd

Where,

f=force

d = distance

Now, $f = 650 + 8x$ lbs,

where x is current amount of cable in shaft

Then,

$dW = fdx\\ = (650+8x)dx$

And x ranges from 0 to 600

Now, the total work is done is

$W =\int\limits^{600}_0 {x} \, (650+8x)dx\\W =\int\limits^{600}_0 \, (650x+8x^{2} )dx\\W = (650\times600 + 4\times(600)^2) \\= 1830000 ft.lb\\W = 1830000 ft.lb$

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