A cable that weighs 8 lb/ft is used to lift 900 lb of coal up a mine shaft 650 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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zainsubhani zainsubhani · Answer 1 · 2020-02-13T09:00:43+01:00

Answer:

Total Work=2275000 ft-lb

Explanation:

According to Riemann sum approximate for work needed to lift the cable:

[tex]W= \lim_{n \to \infty} \sum_{i=1}^n8x_i\Delta x[/tex]

Sine we have to add 650 terms because distance 650 ft, we will us the integration.

[tex]W=\int\limits^a_b {f(x)} \, dx \\W=\int\limits^{650}_0 {8x} \, dx \\W=[4x^2]_0^{650}\\W=4(650)^2-0^2\\W=4*422500 ft-lb\\W=1690000 ft-lb[/tex]

Work done on lifting:

[tex]W_1=900*650\\W_1=585000 ft-lb[/tex]

Total Work= [tex]W+W_1[/tex]

Total Work=1690000+585000

Total Work=2275000 ft-lb

Cricetus Cricetus · Answer 2 · 2022-02-04T15:20:06+01:00

As per the Riemann sum,

→ [tex]W = lim_{n \rightarrow \infty} \sum^n_{i=1} 8x_i \Delta x[/tex]

By applying integration,

→ [tex]W = \int\limits^a_b {f(x)} \, dx[/tex]

[tex]W = \int\limits^{650}_b {8x} \, dx[/tex]

[tex]= [4x^2]^{650}_0[/tex]

[tex]= 4(650)^2-0^2[/tex]

[tex]= 4\times 422500[/tex]

[tex]= 1690000 \ ft-lb[/tex]

On lifting, the work done be;

→ [tex]W_1 = 900\times 650[/tex]

[tex]= 585000 \ ft-lb[/tex]

hence,

The total work done will be:

= [tex]W+W_1[/tex]

By putting the values,

= [tex]1690000+585000[/tex]

= [tex]2275000 \ ft-lb[/tex]

Thus the approach above is right.

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