Finding the Work done

[heykimmyt]

How much work is done to bring a load of ore to the surface if a miner uses a cable weighing 2 lb/ft to haul a 100 lb bucket of ore up a mineshaft 800 feet deep?

ok so... any pointers on where to start?? i began using related rates techniques but that didnt account for the weight and such..

also this problem was given as a "challenge" problem so it's really got me confused..here are the teacher's instructions..

The base of a certain solid is a plane in region R enclosed by the x-axis and the curve y=1-x[sup:273p888g]2[/sup:273p888g]. Each cross-section of the solid perpendicular to the y-axis is an equilateral triangle with its base lying in R. Find the volume of the solid.

 

[heykimmyt]

lol -crosses fingers-

 

[BigGlenntheHeavy]

\(\displaystyle For \ the \ second \ one: \ V \ = \ \sqrt3 \int_{0}^{1}(1-y)dy \ = \ \frac{\sqrt3}{2} \ cu. \ units\)

\(\displaystyle Observe \ the \ graph \ (y \ = \ 1-x^{2}), \ can \ you \ figure \ it \ out?\)

[attachment=0:1vcwzt9u]xyz.jpg[/attachment:1vcwzt9u]

\(\displaystyle What \ would \ be \ the \ volume \ if \ (base \ being \ the \ same) \ the \ cross \ sections \ of \ the \ equilateral\)

\(\displaystyle \ triangles \ were \ parallel \ to \ the \ y \ axis \ instead \ of \ being \ perpendicular?\)

\(\displaystyle Answer: \ V \ = \ \frac{\sqrt3}{2}\int_{0}^{1}(1-x^{2})^{2}dx \ = \ \frac{4\sqrt3}{15} \ cu. \ units.\)

 

[wjm11]

How much work is done to bring a load of ore to the surface if a miner uses a cable weighing 2 lb/ft to haul a 100 lb bucket of ore up a mineshaft 800 feet deep?

ok so... any pointers on where to start?? i began using related rates techniques but that didnt account for the weight and such..

Are you sure it is just a 100 pound bucket? That seems pretty small.

The work has two components: the work done on the 100 lb (45.3592 kg) bucket and the work done on the cable. The work on the bucket is very straightforward: Work = Force x Distance,

Work = (100 lb)(800 ft) = 8000 lb-ft

Or in metric units

Work = mgh = (45.3592 kg)(9.81 m/s^2)( 243.84 m) = 108.5 kilojoules.

You can use related rates on the cable, noting that the cable starts with a weight of

Weight = (length)(weight per foot) = (800 ft)(2 lb/ft) = 1600 lb

and decreases to zero when the bucket reaches the surface. The non-calculus approach is to recognize that the center of mass of the cable is at 400 feet of depth, so the mass of the cable is raised 400 ft.

Work = (1600 lb)(400 ft) = 640000 lb-ft

Now, you do the related rates version.