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Power problem, to use net force or work energy theorem?

  1. Jan 17, 2013 #1
    1. The problem statement, all variables and given/known data

    A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising the helicopter. An 720-kg helicopter rises from rest to a speed of 7.5 m/s in a time of 5.0 s. During this time it climbs to a height of 7.9 m. What is the average power generated by the lifting force?

    2. Relevant equations

    Gpe=mgh KE=1/2MV^2

    3. The attempt at a solution
    Ok so we had a test today and afterwards we were discussing how to do one of the problems. Me and a classmate had two different ways to do this problem. He determined the net force acting on the helicopter by calculating the average acceleration and from there using the equation W=Fcos*X. I used conservation of energy to determine that the total energy in the problem using Gpe=mgh KE=1/2MV^2. From there we both divided by time to get teh power, not the work. Using these two methods does end up with two relatively different answers, so I was wondering which is the right method to use?
     
  2. jcsd
  3. Jan 17, 2013 #2

    TSny

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    Hi deblimp. Welcome to PF!

    The work-energy approach should give the correct answer.

    Why isn't the other method valid? Note that the formula W = F*x*cosθ requires using the force averaged over the distance x. But when you use a = Δv/Δt for the average acceleration, you get the acceleration averaged over time from which you can get the force averaged over time. But the time-average of the force doesn't necessarily equal the distance-average of the force.

    For example, suppose you push a block along a frictionless surface a total distance of 2 meters such that during the first meter you push with 10 N of force and the second meter you push with 20 N of force? The distance-average of the force would clearly be 15 N. Would the time-averaged force be greater than, less than, or equal to the distance average?
     
  4. Jan 17, 2013 #3

    PhanthomJay

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    While both methods should give the same result, you are looking for the average power generated by the lifting force. It appears from your wording that your classmate was using the net force to determine the net work done, then dividing the net work done by the time to get the average power. This would be incorrect. I don't know what energy equation you were using either, for that matter. You might want to show your work, and that of your classmate also, if you have it.
     
  5. Jan 18, 2013 #4

    CWatters

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    How did he calculate the acceleration?

    Method 1 (using the height and velocity data):

    V2 = U2 + 2as

    a = V2/2s = 7.52/(2 *7.9) = 3.5ms-2

    Method 2 (using the height and time data):

    s = ut + 0.5at2
    a = 2s/t2 = (2 * 7.9)/52 = 0.53ms-2

    Different answers suggests the acceleration wasn't constant and that's a requirement for using these equations. He made a wrong assumption I believe.

    Your approach using energy is correct.
     
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