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Challenge problem - Power

  1. Feb 8, 2013 #1
    A force F is acting on an object whose mass is 1kg. The force in newtons is 4+4t², where t is the time. The angle in radians that the force does with the displacement is 2πt. If the object was initially at rest, estimate the power due to that force at t=3s.

    I've tried to solve it but I've failed.
    Here is what I got until now

    P=power
    v=velocity
    m=mass
    F=force
    θ=angle force does with displacement
    x=displacement
    W=work

    P = dW/dt = d([itex]\vec{F}\bullet\vec{x})/dt = \vec{F} \bullet \vec{v} + d\vec{F}/dt\bullet\vec{x} [/itex]


    W = [itex] \int \vec{F}\bullet\ d \vec{x} = \int F.dx.cos(\theta) [/itex]
    W = mv2/2

    I don't know how to go on from here.
     
  2. jcsd
  3. Feb 8, 2013 #2

    TSny

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    The correct expression for the work done during an infinitesimal displacement is ##dW = \vec{F}\cdot\vec{dx}##

    So, what expression do you get for the power ##P = dW/dt##?

    Does the object move along a straight line? Or in 2 or 3 dimensions? [EDIT: Nevermind, I think it must be moving in 2 or 3 dimensions and F is the only force.]
     
    Last edited: Feb 8, 2013
  4. Feb 8, 2013 #3
    So the dF/dt . x goes out?


    2 dimensions
     
  5. Feb 8, 2013 #4

    TSny

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    Yes.
     
  6. Feb 8, 2013 #5

    I like Serena

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    Did you try to find an expression for the acceleration in the direction of the velocity?
     
  7. Feb 8, 2013 #6
    Actually I've solved it considering the "displacement" term was instantaneous displacement (that has the direction of velocity). But I think displacement was used to design the vector that points from the initial position to the actual position of the object. In the first case:

    a=Fcos(θ)/m = (4+4t²)cos(2πt)

    Velocity is v = ∫a.dt = 6/π2
    So P=F.v.cos(θ) = F.v = 240/π2

    Is this right? What would be the result in the second case?
     
  8. Feb 8, 2013 #7

    I like Serena

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    Yes. That is right!

    Displacement is the change in position in a (small) time interval.
    It is always in the direction of the velocity.
     
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