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Calculate the lift work?

  1. Jan 30, 2015 #1
    1. The problem statement, all variables and given/known data
    A crane lifts a load of 3000 kg to a height of 8.5 meters in 15 seconds.

    Calculate the lift work?

    2. Relevant equations


    3. The attempt at a solution
    This is a very simple problem but the thing is its unclear data. The given data does not clarify whether or not this is an accelerated motion. So there will be two ways to solve it.


    1. The first solution

    Assuming that this is a uniform motion.

    F=P=m.g
    A= F.s = m.g.h
    = 3000.(9.81).(8.5)= 250,155 J

    2.The second solution

    Assuming that this is an accelerated motion

    F-P = m.a → F= (a+g)m

    a= 2h/t2 = 2.(8.5)/152 ≈ 0.076 m/s2

    →F= (0.076+9.81).3000=29658 N → A=F.s=29658.(8.5)= 252,093 J

    I want to ask you that how can we know whether or not this is an accelerated motion so that we have only one solution. Thanks
     
  2. jcsd
  3. Jan 30, 2015 #2

    haruspex

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    Do you understand why your accelerated solution gives a larger number? What has happened to the extra energy? Are these the only two possibilities?
     
  4. Jan 30, 2015 #3
    That's not my question. My question is how can we know whether this is a uniform motion or an accelerated motion based on the limited given data only. I must know that because this problem is taken from a test, therefore one single solution is allowed.
     
  5. Jan 30, 2015 #4

    haruspex

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    the answers to my questions resolve your question.
     
  6. Jan 30, 2015 #5
    Haruspex. As a science advisor, could you figure it out whether this is an accelerated motion or a uniform motion based only on the given data.
     
  7. Jan 30, 2015 #6

    haruspex

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    It can be accelerated yet yield the lower number. You are assuming uniform acceleration.
     
  8. Jan 30, 2015 #7
    Alright. your question is:
    why your accelerated solution gives a larger number?
    Because you have to add a to F= (a+g)m

    What has happened to the extra energy?
    You need extra energy in order to accelerate it. That why you get more energy in the second solution.

    Are these the only two possibilities?
    There is another possibility. That is -a, meaning that the object is slowing down when it is reaching the height of 8.5 meters. Therefore the energy will be smaller then the two first possibilities.

    But answering to these question does not satisfy my question.
     
  9. Jan 30, 2015 #8
    Haruspex, we are not making progress here. you know what I mean.
     
  10. Jan 30, 2015 #9
    Hellô. Is there any body here to help me out ?
     
  11. Jan 30, 2015 #10

    haruspex

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    None of those were quite the answers I was looking for.
    Another possibility is that the acceleration changes. It could accelerate from rest, then decelerate as it approaches the required height.
    The extra energy in your constant acceleration model ends up as KE because you have it still moving.
    The question does not require it to be still moving. The question (as you've stated it) does not even make clear whether it is initially moving (which you have to assume for your uniform speed solution).
    So you see, there are all sorts of possible answers, and in the end you have to make some reasonable assumption. I can see these main options.
    1. It is initially moving at the required speed and still moving at the end of it.
    2. It is initially at rest. The residual KE is to be minimised. The fastest possible deceleration is g.
     
  12. Jan 30, 2015 #11
    Are you implying that the person who made this test is not completely understand the problem and what's why he or she provides a very vague data which is not enough to solve it with one single solution ?
     
  13. Jan 30, 2015 #12

    haruspex

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    There's no doubt that (if that is the exact wording) there is not enough information. However, it turns out that both the main options I suggested produce the same answer, so I'm reasonably certain that is the intended answer.
     
  14. Jan 30, 2015 #13

    BvU

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    This the same test as the car bulbs exercise ? Looks to me as a teacher (or committee) that wants to add some real life aspects to otherwise dry exercises. So they let you guess some needed things and they spice in some noise (15 sec, provided you make the obvious assumption). The wording "lift work", in combination with the absence of any other useful info would be my clueue ( :) ) to do potential energy only. Without hesitation if there is no further context. With the risk of being wrong.
     
  15. Jan 30, 2015 #14

    ehild

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    Assume the load is in rest. What is the reasonable way to lift it to a given height? It must be accelerated to some speed otherwise it will not move upward. But it is not practical if the load moves when it reaches the height. It must stop there.

    So the crane accelerates the load for some time and then lets it go up "by itself" . The load has to loose its speed just at the given height.
    You can assume constant acceleration during the first stage. Determine the work done by the crate. The work will not depend on the details of the lifting process if the final speed is zero. Which is not surprising, as the increase of potential energy is equal to the work done on the load.
     
  16. Jan 30, 2015 #15
    in short, you mean that the first solution W=mgh is more suitable or something?
    The load of course 'must' start at rest. But when it reached the height of 8.5 meters, we don't know whether or not it's still moving. We cannot say that it 'must' stop at the height of 8.5 meters.
     
  17. Jan 30, 2015 #16

    ehild

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    I said that it is the best way to lift a load if it does not move when it reaches the desired height. That means it is enough to rise it to a lower height and then let it go by its inertia and decelerated by gravity.
    In this case, accelerating the load to 1.13 m/s speed, it is enough to lift up to 8.43 m instead of 8.50 m. Then it reaches the height with zero velocity in 15 s.
     
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