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Elevator falling on springs

  1. Aug 3, 2015 #1
    1. The problem statement, all variables and given/known data
    My question states that an elevator is falling on springs that are supposed to absorb the residual energy if the brakes fail to stop the elevator. They are asking how many springs are needed so that the elevator does not compress the springs by more than 0.2m?
    mass of elevator is 4000 kg
    the distance the elevator falls is 15 m
    the sum of frictional forces of the breaks is 5000N
    and the spring constant of each spring is 5.0x10^6 N/m with a maximum compression of 0.3 m

    2. Relevant equations

    3. The attempt at a solution

    I found the work done by the frictional forces by multiplying the frictional forces by the distance, and subtracted that by the potential energy of the elevator so obtain the total energy of the elevator.
    This gave me:
    Wf=(Ff)(d)=(-5000N)(15m)= -75,000 J
    Ep =mgh=(4000kg)(9.8N/m)(15m)= 588,000 J
    Etotal= 513,000 J

    I then found the elastic potential energy of each spring by using the compression of 0.2 m
    Ep=(0.5)(5.0x10^6N/m)(0.2m^2)= 100,000 J

    To find how many springs I need, would I just have to divide the total energy of the elevator by the potential energy of the springs? I'm not sure what the next step should be
     
  2. jcsd
  3. Aug 3, 2015 #2
    When calculating the change in gravitational potential energy of the elevator,wouldn't it be more appropriate to consider 'h' as 15.2 rather than 15?
     
  4. Aug 3, 2015 #3

    mfb

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    That is possible, right.

    Such a small braking distance won't help passengers, by the way, it still corresponds to a deceleration in excess of 50 g.
     
  5. Aug 3, 2015 #4
    Possibl
    Thanks!
     
  6. Aug 4, 2015 #5

    haruspex

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    And it asks if the brakes fail to stop it?!
    Maybe it means that failing brakes can still be assumed to achieve 5000N?
     
  7. Aug 4, 2015 #6

    mfb

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    The brakes fail to stop it completely, but they still work and exert a force (apparently not with their normal force, because normally they have to hold the elevator, assuming the counterweight is counted as "brake"). That's how I interpreted the problem.
    Otherwise it would not make sense to discuss brakes.
     
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