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Work and Kinetic Energy

  1. Mar 3, 2008 #1
    Having trouble with the following two part problem:

    A) Joe, mass 89.7 kg, is racing against Tom. When Joe and Tom have the same kinetic energy, Tom is running faster. When Joe increases his speed by 22.8%, they are running at the same speed. Find Tom's mass.

    Tried equating their kinetic energies and solving for the unknown mass, but that did not work. The reasoning behind it seemed logical, I was surprised that the answer turned out to be wrong.

    B) A highway goes up a hill, rising at a constant rate of 1.00 m for every 50 m along the road. A truck climbs this hill at constant speed vup = 16 m/s, against a resisting force (friction) equal to 1/25 of the weight of the truck. Now the truck comes down the hill, using the same power as it did going up. Find vdown, the constant speed with which the truck comes down the hill.
    ASSUME: the resisting force (friction) has the same magnitude going up as going down.

    This one is really tough, probably the hardest all semester. All I know is the slope of the highway going up, 1/50, and that the coefficient of friction is a fraction of the force needed to move the car at constant v.

    I would appreciate any guidance on this one, to steer my thoughts in the right direction. I am not just looking for solutions here.
     
  2. jcsd
  3. Mar 3, 2008 #2
    A) Can you show your work for A? I think you're on the right track with setting their kinetic energy equal, (and then using what you know about Joe's velocity in terms of Tom's velocity) but I'm not sure how your method didn't work.
     
  4. Mar 3, 2008 #3
    1/2 mv^2 = 1/2 mv^2
    1/2 m (.228v)^2 =1/2 mv^2
    1/2 m .052v^2 = 1/2 mv^2
    .052 (89.7) = m
     
  5. Mar 3, 2008 #4

    jambaugh

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    destro47. You'll need to distinguish their masses and velocities:
    e.g. 1/2 M1 V1^2 = 1/2 M2 V2^2....
     
  6. Mar 3, 2008 #5
    the left side of the equation is Joe and the right is Tom
     
  7. Mar 3, 2008 #6
    Since Joe has to increase his velocity by 22.8% to match speed with Tom, you want 1.228vJoe = vTom, not .228vJoe=vTom.
     
  8. Mar 4, 2008 #7

    jambaugh

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    Ok but using sides of the equation to distinguish variables is very very very bad notation. You can't then apply any of the usual algebraic techniques. If you want to avoid subscripts let one of two's variables be represented with capital letters.
     
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