Rope question - integration

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You are correct that the rope weighs 2 lb/ft, so 50 ft of rope would weigh 100 lb.

After t seconds, the rope has been lifted t ft, so the length of rope being held will be 100-t ft. Therefore $F(t)=2(100-t)+50$, since the rope weighs 2(100-t) lb and the bucket weighs 50 lb.

To find the total work done, you have to integrate $F(t)$ from $t=0$ to $t=100$.

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Updated on July 27, 2020

Comments

  • Panthy
    Panthy over 3 years

    A 50-lb bucket is at the bottom of a 100-ft well. A 200 lb rope (also 100 ft long) is tied securely to the bucket. We will use rope to lift this bucket out of the wall, at a rate of 1 foot every second.

    (a) How much does this rope weigh per foot? How much would a 50 foot length of this rope weigh?

    (b) The force that is applied to the system, F(t) at a given time t is equal to the total weight that you are holding in the well: the weight of the rope remaining in the well, plus the weight of the bucket. What is F(0) F(50), F(100)? Find a formula F(t) that describes the force at time t.

    (c) Find the total work done to lift the bucket to the top of the well.

    I think I know how to do the first part but not sure so I'm just gonna say what I did:

    Since the rope is 100 ft long and it weighs 200 lb's, shouldnt its weight per foot just be 200/100 = 2 lb/ft? If thats right then it should be 4 lb/ft for a 50 ft length rope.

    For part b im lost and I'm pretty sure I need part b done to be able to do part c.

    Help is required,

    THanks

  • Panthy
    Panthy over 9 years
    I understand what you said except I don't get why its 2(100-t) instead of just (100-t)
  • user84413
    user84413 over 9 years
    The length in feet is 100-t, but the rope weighs 2 lb/ft, as you said, so its weight is 2(100-t).
  • Panthy
    Panthy over 9 years
    thanks :) appreciate your help