A cable that weighs 6 lb/ft is used to lift 700 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.)

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Answer:

15000000ft-Ib

Step-by-step explanation:

We were told to Show how to approximate the required work by a Riemann sum which is a certain kind of approximation of an integral by a finite sum.

. So we will need to divide the cable into 6 segments,

✓let us denote x as the distance between the top of the mine shaft and the segment.

✓Let us denote ∆x as the length of the segment

Then work done on a segment, work done on the coal and total work done can now be calculated.

CHECK THE ATTACHMENT FOR THE DETAILED EXPLANATION

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Suppose that the weight of a cable is [tex]6 \ \frac{lb}{ft}[/tex] and used to lift [tex]700\ lb[/tex] of coal up a mining shaft [tex]600\ ft[/tex] deep.

  • Let the ith portion of a cable have a width of [tex]\Delta x[/tex], therefore the weight of the ith part of the cable is [tex]6\ \Delta x[/tex], and the distance of the ith part of the cable from the mineshaft is [tex]x_i^{*}[/tex].
  • As a result, the effort done on the ith segment of the wire is [tex]6x_i^{*}, \Delta x[/tex]. The overall amount of work completed for the cable is as follows:

[tex]\to W_{ca} = \lim_{n\to \infty} \Sigma^{n}_{i=1} 6x_i^{*} \Delta x \\\\[/tex]

           [tex]=\int^{600}_{0} 6x \ dx\\\\= [6x^2]^{600}_{0} \\\\= 3 (360000)\\\\ = 1080000\ ft-lb\\\\[/tex]

Thus,  the work done again for cable is [tex]W_{ca} = 1080000 \ ft-lb[/tex].  

If the cable weighs 700 lbs and the distance is 600 feet, As a result, the effort expended in raising the coal to the top of the mineshaft is referred to as

[tex]\to W_{co} = 700 \times 600 = 420,000 \ ft-lb[/tex]

As a result, the total amount of work completed is as follows:

[tex]\to W =W_{ca}+W_{co} \\\\[/tex]

         [tex]= 1080000+ 420000 \\\\= 1500000 \ ft-lb \\\\[/tex]

Hence, the total work done is [tex]\bold{W = 1500000\ ft- lb}[/tex].

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