How could you determine the amount of work required to lift a given filled container?

Question

A cable weighs 2 lb/ft is attached to a bucket filled with coal that weighs 800 lbs. The bucket is initially at the bottom of a 500-ft mine shaft. Determine the amount of work required to lift the bucket to a) the midpoint of the shaft, b) from the midpoint to the top of the shaft and c) all the way up to the shaft.

Answer 1

this problem REQUIRES integral calculus because the weight of the rope is constantly changing as its length decreases as it is being pulled up, you can't do it with quadratics nor can you average out weights because the graph of the work done throughout the process is a curve so between the midpoint (250 ft) and one of the 2 ends (0ft, 500ft) the rate of change is slower in beginning resulting in higher work done at beginning of travel then at the later half so rate of work done is NOT a constant change

We know that Work = Force* Distance

so we know the force will be changing so that will be our integrand F(x)

now calculate the force

F(x)= weight of bucket + weight of rope

F(x)= 800+2(500-x)

F(x)= 1800-2x

now setup the integral equation for each part a, b, c

a) W= ∫ (b= 0) (a= 250) 1800-2x dx

W= 1800x- x^2 | b= 250 a= 0

W= 1800*250-250^2

W= 387500 ft/lb

b) W= ∫ (b= 250) (a= 500) 1800-2x dx

W= 1800x- x^2 | b= 500 a= 250

W= (1800(500)-500^2)-(1800*250-250^2)

W= 262500 ft/lb

c) W= ∫ (b= 500) (a= 0) 1800-2x dx

W= 1800x- x^2 | b= 500 a= 0

W= 1800*500-500^2

W= 650000 ft/lb

sorry for the weird way of writing the integrals, couldn't find a way to convert it to proper notation

anways hope this helps

Answer 2

work = force * distance

First of all, I'll assume you are pulling the cable straight up, not with a pulley.  If you are using a pulley, then the weight of the hanging cable is constantly changing.

In part (a), you are pulling the coal 250 ft up.  This will require enough force to pull both the coal AND the cable.  The coal is 800lbs.  The cable is 2lb/ft and is extending 500ft down the shaft, so the weight of the cable is: 2 * 500 = 1000 lbs.  In total, you are pulling 1800 lbs.

Now I will convert to newtons and meters so our answer is in Joules, the standard unit of energy (work).  250 feet is 76.2 meters.  1800 pounds is 8006.8 newtons.

So we have:

a) w = fd = (8006.8)(76.2) = 610,118.16 J

b) same distance, but less cable is required (assuming you are attaching the cable from midway and aren't just continuing from before).  The cable's weight is: 250 * 2 = 500 lbs.  In newtons, this is 2224.1.  So your answer becomes: (2224.1)(76.2) = 169,476.4 J

However, if you're just continuing to pull it up from before, then the answer is the same as (a).  The problem is leaving out some key details...

c) If you start from the bottom and go straight to the top without reattaching the cable, then the answer is: (8006.8 N) * (152.4 m) = 1,220,236.32 J

These answers seem very high for a regular textbook problem.

If you ARE using a pulley and the cable's weight does change, then here's what you do for that situation.

You could use calculus, but that's not necessary.  You could use a quadratic function, but that's also not necessary.  All you have to do is average out the weight of the cable.  Just add: (weight at the bottom) + (weight at the top).  Then divide this sum by 2, and there's your average weight for the trip up.

So for part (a), here's how we'd find the average cable weight.  At the bottom, it's extended 500 ft, so that's 1000 lbs.  At the top of the trip (midway up the shaft), it's extended 250 ft, so that's 500 lbs.  At those together and you get 1500 lbs.  Divide by 2 and you get 750lbs, and that's the average.  Convert to N and you get 3,336.2

So (a) becomes: (3336.2 N) * (76.2 m) = 254,218.4 J

for b: average weight = (500lbs + 0) / 2  =  250 lbs  (at the top it has no length and therefore no weight, that's where the 0 comes from)

250lbs is 1112.1 newtons

b) 1112.1 * 76.2 = 84,742 J

for C, the average weight is simply: (1000 + 0) / 2  = 500 lbs, which is 2224.1 newtons

c) 2224.1 * 152.4 = 338,952.8 J

So now, no matter what your teacher or the book was trying to ask, you have the right answers.