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Find work given mass, acceleration, and time.

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Poll closed Apr 3, 2012.
  1. yes

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  1. Mar 30, 2012 #1
    1. The problem statement, all variables and given/known data

    A 510 kg, light-weight helicopter ascends from the ground with an acceleration of 2.30 m/s^2.
    Over a 5.50 s interval, what is the work done by the lifting force?

    a=2.3m/s^2
    m=510kg
    t=5.5s




    2. Relevant equations


    f=m(a)
    d=.5(a)t^2
    W=F(Cosθ)*d



    3. The attempt at a solution

    Find force = m(a) = 510kg*2.3m/s^2 = 1173N

    Find distance = .5(a)t^2 = 34.7875m

    use this force and distance in w=f(d) to find work (in Joules):

    1173*34.7875 = w
    w = 40805.74J
     
  2. jcsd
  3. Mar 30, 2012 #2
    Ok this is fairly simple, if the helicopter travels at a speed of 2.3m/s^2 and it traveled for 5.5 seconds, then how many meters did it travel in that 5.5 seconds? That is the distance.
     
  4. Mar 30, 2012 #3
    I'm not going to ask you to do it for me, but using the equation above, I already solved for distance. Unless that is the wrong equation.
     
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