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B Energy required to slow mass falling from "X" height

  1. Jun 14, 2017 #1
    I'm trying to figure out how much energy needs to be absorbed, radiated or displaced from an object falling from various heights. To make this clear, the "item" is a person so the weight can vary from 70-350lbs and the height let's say 60-3500 ft. Obviously the speed at the end will determine a lot and needs to be reasonable (IDK what that is, maybe 2-3m/s just to to start with).

    In the scenario I'm looking at, I think the fall rate needs to be constant or somehow decreased the last 10-15m to an acceptable speed

    So if the height is 1000m and weight is 100kg and descent speed is a constant 10m/s, the person needs to be slowed to 2 or 3m/s over the last 15m. I'm trying to come up with a formula to calculate the energy generated from the first 985m and how much at the last 15m. I'd like to be able to swap out number to play around and see how the numbers impact energy generated/liberated.

    I need to calculate how much heat would be generated by the 985m descent as friction would be used to keep it from accelerating - then I need to calculate the same for the last 15m.

    Can someone help with finding the proper calculations?
  2. jcsd
  3. Jun 14, 2017 #2


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    Staff: Mentor

    IF they fall from far enough, they will be at "terminal velocity". Look that up on Wikipedia to get typical values.

    Then their kinetic energy is just KE = 1/2 m*v^2, so you can calculate that (using appropriate units). Slowing them to a stop will require that amount of energy.
  4. Jun 14, 2017 #3


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    Staff: Mentor

    Are you ignoring air resistance? If not, that's going to complicate things immensely.

    Assuming air resistance is negligible, then this is easy relatively easy. Find the potential energy of the person at height ##H##. Then, using the final velocity you want, find the kinetic energy of the person. Subtract the kinetic energy from the potential energy and you have the total energy you need to expend.

    The only way for the descent speed to remain constant is to expend energy somehow. A person of 100kg requires 981 newtons of force to remain at a steady velocity against gravity, so the energy expended during this part is equal to the work performed by that force: ##W=Fd##, where ##d## is the distance the person falls while under the influence of this force.

    Subtract the work done during the descent and the final kinetic energy from the original potential energy. This should be how much more energy you need to expend.

    Assuming friction is the only force acting on the person, then the energy lost as heat is equal to the work done by friction during both parts of the descent.
    If friction isn't the only force then you'll have to account for the work done by any other forces.
  5. Jun 14, 2017 #4
    Thanks for the reply!

    I am not factoring in air resistance. Maybe this will make things more clear. Someone hooked on a spool of line/cable that is released as they descend. The spool is attached to either an induction motor (or generator of some kind) or a brake pad/clutch plate of some kind. If a motor is used, electricity is dumped into a resistor(s) bank as heat or the brake pad/clutch creates friction heat.

    So what I'm seeing here is that as they fall the 985m, it will be a constant amount of energy dissipated (during the length of the descent) as the speed and weight remain the same. The last 15m a greater amount of energy will need to be dumped as it slows to the final rate.

    PE of 1000m, 100kg = 980,665 joules
    So PE for 985m, 100Kg = 965,300 joules
    If falling at 10m/sec then 985/10 = 98.5 seconds
    965,300/98.5 seconds = 9,800 joules per second

    Final speed KE values
    2m/s = 200
    3m/s = 450
    4m/s = 800
    5m/s =1250
    6m/s =1800
    7m/s= 2450
    8m/s =3200
    9m/s =4050
    10m/s =5000

    I'm not certain about your post which I bolded. Are you saying that 981 newtons is needed to counteract gravity so as not to increase/accelerate the speed of fall? That is what makes sense to me. I can't see that additional energy needed to help something fall...
  6. Jun 14, 2017 #5


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    Staff: Mentor

    Yes, that's what I'm saying. Applying 981 newtons upwards on the person ensures that no acceleration (or deceleration) occurs.
  7. Jun 15, 2017 #6
    I think, in the end, the total amount of heat generated will be equal to the amount of gravitational potential energy lost. It will be generated at a rate proportional to the rate of descent, so slightly less at the top where the person is accelerating to constant descent speed, and slightly more at the bottom where you have to bring the person to a near stop. An automobile disc brake might work well.
  8. Jun 15, 2017 #7


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    Since there will be 985 meters of cable extended at the end of the of the fall, then the elasticity of the cable needs to be considered in the chosen deceleration rate or there can be some pretty severe oscillations at the bottom. (Off main topic but relevant to intended application)
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