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Homework Help: Work done on a lift, help please!

  1. Apr 17, 2009 #1
    1. The problem statement, all variables and given/known data
    a block of mass m = 20 kg is placed on the floor of a lift (mass M = 1000 kg), the lift is being pulled up distance d =25m by a cable, the normal force on the block FN has a constant magnitude of 300N, how much work is done on the lift by the force on the cable?

    2. Relevant equations

    Wgravity = -mgd m=mass, d = distance, g= 9.8

    W = Fd

    Wnet = [tex]\sum Work[/tex]

    3. The attempt at a solution

    For Normal force I used W = Fd >> 300N x 25m = 7500J

    For Wgravity I did >> -(20+1000) x 9.8 x 25 = -249900 J

    Then Wnet = -Wgravity + W = -249900 + 7500 = -242400 J or -242.4 KJ

    for some reason I don't think this is right? can anybody give me any advice please? I have not been provided with the actual answers.
  2. jcsd
  3. Apr 17, 2009 #2


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    The work apparently resulted in an increase in potential energy all right. But it also apparently resulted in an increase in kinetic energy as well.

    At 25m what is the speed of the elevator?
  4. Apr 17, 2009 #3
    lowlypion there is no speed given in the question. if I work it out then would it be V = [tex]\sqrt{2K / m}[/tex] , what value would I use for Kinetic energy? Wnet or W?
  5. Apr 17, 2009 #4
    Hi, since LowlyPion is offline, I'll give You a clue:
    the lift is being pulled up from initial velocity 0, now it had to have speed after X meters,
    how can You find acceleration with a mass of a body and the normal force exerted by the floor on the body?and than you should find the velocity.
    good luck
  6. Apr 17, 2009 #5


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    Actually you are right, in that they don't give the initial speed and so that is not a good way to go. My comment was really directed at the notion that the work against gravity not relating to the total energy should be a clue that work was going into increasing kinetic energy.

    So look at it another way.

    If the force on the block is 300N then the force on the cable will be in the same ratio.

    Hence I think you can say

    300N / 20 kg = Fc / (1000 + 20)

    Fc = 15*(1020)

    So work by the cable over the same distance should look like

    W = Fc * d
  7. Apr 17, 2009 #6
    Just wanted to approve the answer.
    done in a different way and got the same answer.
    If you're interested:
    m=20kg weigh in non accelerating lift = 200 N , in this one 300N thus acceleration = 5m/s^2
    using kinematics V^2=250
    W= change in Ke and Ep= m*250/2 + m*250. while m=1020.
    same answer! awesome :D
    IMO LowlyPion's method is better, although i didn't quite understood it...
  8. Apr 17, 2009 #7


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    Sure you do.

    F = m*a

    Acceleration is (must be) the same.
  9. Apr 17, 2009 #8
    awww , awesome ! stupid me XD
  10. Apr 18, 2009 #9
    hmm I can see what you are saying that the two forces can be the same, but I dont quite understand why the Work gravity does not affect it? since the lift is moving upwards, does gravity not affect it? sorry if its a stupid question. thanks two both of you guys for your help.
  11. Apr 18, 2009 #10


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    Of course gravity enters into it. And some of the work of the lift goes into raising the mass over the 25m. But the statement of the problem indicates that there is an excess of energy developed, that will go into kinetic energy.

    Looking at the weight of the block on the floor you can see that the 20 kg times 9.8 is short of the 300N force given from the motion. So part of the 300N is gravity induced (196N of it ) and the remainder ( 104N ) goes into acceleration of the block.

    Well if the block is accelerating then so is the lift. And if the lift is accelerating then the lift's mass acceleration plus its weight at rest, will be in the same proportion as the blocks acceleration and its weight at rest.
  12. Apr 18, 2009 #11
    Omg I never saw it like that, how'd you get so smart lol. Thanks alot lowlypion.
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