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Please help! Work,power and energy question

  1. Sep 9, 2012 #1
    :( Please help! Work,power and energy question

    1. The problem statement, all variables and given/known data
    There is a 100kg object on a 30-degree incline ramp.
    The mew of friction is 0.5.
    There is a motor on top of the ramp, which accelerates the object at .22 m/s^2 by a rope.
    Find the power output of the motor.


    2. Relevant equations
    d=(vit)+(.5)(a)(t)^2
    W=Fd
    F=ma
    P=w/t, E/s


    3. The attempt at a solution
    I think I'm supposed to use Ek+Ep+TE=0,
    but how am I supposed to find the power without knowing the time?
    So, I've tried this method:

    P=J/s
    J(work)= Fxd
    So I thought maybe I'll use t=1:

    vi=0
    vf=x
    a= 0.22
    d=?
    t=1

    d=vit+(.5)(a)(t)^2
    d=(.5)(.22)(1)
    d=.11m

    Fnet=ma
    Fa-Ff=ma
    Fa-(100kg)(9.81)(cos30)(.5[mew])-(100kg)(9.81)(sin30)=(100)(.22)
    Fa-424.79-490.5=22
    Fa=937.29N

    W=Fd
    W=(937.29)(.11)
    W=103.10J

    P=J/s
    P=103.10Watts

    So here's the one approach I had,
    ------------------
    Here is the second

    Ek+Ep+TE=0
    (.5)(m)(vf^2-vi^2) + mgh + (Ff)d=0

    To find Vf,
    Vi=0
    Vf=?
    a=0.22
    d=x
    t=1s

    a=(Vf-Vi)/t
    .22=Vf/1
    Vf=.22 m/s

    plugging it back in,
    (.5)(100kg)(0.0484) + (100)(9.81)(h) + (100)(9.81)(cos30)(0.5[mew])d=0

    To find d,
    Vi=0
    Vf=x
    a=.22
    d=?
    t=1s

    d=vit+(.5)(a)(t^2)
    done above,
    d=0.11m

    (.5)(100kg)(0.0484) + (100)(9.81)(h) + (100)(9.81)(cos30)(0.5)(0.11) = 0
    2.42 + 981h + 46.73 = 0
    h=.05m

    Then plug the height of .05m,

    to get 98.30J/s?
    --------------


    Another question:

    If there is an elevator moving downwards to the ground at a constant velocity,
    Is the work done 0J? Is the net work between Ft and Fg the same? How should I approach this question? I know there is NOT an acceleration. W=Fd, W=mad, so no work?
    Although there is a ΔEnergy?

    Another question:

    Momentum question,
    Is the "internal work" done by an explosion equal to the combined kinetic energy of the particles? Or is it 0J? How can i visualize internal work in a bomb-like scenario?


    Thank you very much.
     
  2. jcsd
  3. Sep 9, 2012 #2

    CWatters

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    Re: :( Please help! Work,power and energy question

    1) Normally...

    Energy = force * distance
    so
    Power = force * velocity

    However in this case the object is accelerating so the velocity and power can't be a constant value. It will be a function of time.

    I'd draw the free body diagram and calculate all the forces acting down the slope. Then multiply by the equation for the velocity V(t)=at.

    2) No.

    A stationary elevator is subject to 1g.
    An elevator accelerating upwards at 9.8m/s/s relative to the building is subject to 2g. (One due to gravity and one due to it's motion).
    An elevator descending at a constant velocity is subject to 1g
    An elevator accelerating downwards at 9.8m/s/s is subject to zero g.

    So although not accelerating it is being accelerated. There is a force.
     
  4. Sep 9, 2012 #3

    PhanthomJay

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    Re: :( Please help! Work,power and energy question

    The first question is missing information, so it cannot be solved. On the assumption of the one second duration, your first approach yields the average power delivered by the motor. Your second approach to this problem is muddled..you left out the work done by the rope for one thing.

    Regarding the elevator problem, the term net work refers to the total work done by all forces acting on the elevator. No acceleration implies no net force and no net work. That implies that the work done by gravity and the work done by the cable have the same magnitude but of opposite signs.
     
  5. Sep 9, 2012 #4
    Re: :( Please help! Work,power and energy question

    1)

    E = fd,
    P = fv,

    So I guess I should just use 1s here as well to find Vf?

    Vi=0
    Vf=?
    a=0.22
    d=x
    t=1

    a=(Vf-Vi)/t
    0.22=(Vf)/1
    Vf=0.22m/s

    To find F,
    Fgx= (100kg)(9.81)(sin30)
    Fgx= 490.5N

    Ff= (100kg)(9.81)(cos30)(0.5[mew])
    Ff= 424.79N

    Fa-Ff-Fgx=ma
    Fa-424.79-490.5=(100)(0.22)
    Fa-915.29=22
    Fa=937.29N

    P=fv,
    P=(937.29N)(0.22m/s)
    P=206.2 Watts?

    2) The question about the elevator is a very basic level physics question, doesn't touch G-forces. It's highschool physics, but the one of the higher-level questions.

    The first question is not missing any information, I was thinking how I can solve for the power without knowing the time. It's one of those higher-level questions for the elementary physics problems. What is your opinion on the P=fv approach I took instead of the first approach I had?

    Thank you for clearing up the elevator problem. Even though there is a ΔEp, there is no Net Work done? This is hard for me to understand.

    Thank you.
     
  6. Sep 9, 2012 #5

    PhanthomJay

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    Re: :( Please help! Work,power and energy question

    DoJang:
    You say there is no missing information, but yet, you had to make an assumption that t = 1 second to find the power. Suppose you assumed t = 10 seconds; then you would find v = 2.2 m/s, and average power would be 10 times higher than before (1030 watts) and instantaneous power F(v) would be 2060 watts, also ten times higher. Your assumption is no better than mine; you need more info. BTW, work/time is average power; instantaneous power is F(v). The two are equal only when speed is constant , no acceleration.
    Edit: I fogot to addres your question
    Gravitational potential energy changes are due to the work done by gravity only, a conservative force. There is no potential energy function associated with non conservative forces like rope tensions. When gravity does work, there will be a gravitational potential energy change, regardless of whether work done by other forces results in no net work being done.
     
    Last edited: Sep 9, 2012
  7. Sep 10, 2012 #6

    CWatters

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    Re: :( Please help! Work,power and energy question

    No. That would just give you the power required after t=1s at which point it is some arbitrary distance up the ramp. The velocity is changing so the power is increasing constantly. It will be consuming more power after t=2s.

    As others have said, there is something missing from the problem OR you will have to give the power as a function of time.

    Is the height of the ramp known? If so you could work out the velocity at the top which would give you the maximium power needed.
     
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