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The change in kinetic energy if distance is constant but angle of force changes.

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A toboggan is initially moving at a constant velocity along a snowy horizontal surface where friction is negligible. When a pulling force is applied parallel to the ground over a certain distance, the kinetic energy increases by 47%. By what percentage woud the kinetic energy have changed if the pulling force had been at an angle of 38 degrees above the horizontal?

    2. Relevant equations
    since work is the change in kinetic energy:
    w=(Fcos[tex]\theta[/tex] )([tex]\Delta[/tex]d)

    3. The attempt at a solution

    w=(Fcos[tex]\theta[/tex] )([tex]\Delta[/tex]d)
    0.47 = (Fcos1)(1)
    0.47 = F

    next

    W = (0.47cos38)(1)
    = 0.37

    Is 37% percent change the correct answer? The one shown in my text book is 16%, im pretty sure it wrong unless im doing this all wrong.
     
  2. jcsd
  3. Mar 8, 2009 #2
    You have the equations right, but used percentage wrong. Stop reading now if you want to figure it out yourself.

    Because kinetic energy increases, you should express W = 1.47 instead of W = 0.47. This yields a solution which is >1, which means that the energy really increases. If your solution was the answer, the energy would decrease. Thus, you'll obtain 1.16 as the answer -> 16% increase.
     
  4. Mar 8, 2009 #3
    thanks a lot!
    just wondering when there is a increase in kinetic energy, the answer i will get will be the previous kinetic energy + the increase. Which would be why it is 1.47.
     
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