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Kinetic Energy problem

  1. Oct 26, 2003 #1
    I dont know why I am getting a different answer but here is the question:

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

    A few things:

    Ekf = Final kinetic energy
    Eki = Initial kinetic energy
    vf=final velocity
    vi=initial velocity

    Case 1 (where the force applied is parallel to the ground)

    2Facos0(mvf-mvi)=2(100 x 0.47)

    mvf-mvi=(100 x 0.47)/Fa

    Case 2 (where the force is at 38° to the horizontal)

    2Facos38(mvf-mvi)=2(100 x y)

    mvf-mvi= (100 x y )/FaCos38

    Now equating the two equations:

    100 x 0.47 = (100 x y)/cos38

    Solving for y I got y=0.37

    So the increase is ~ 10%, and the answer is 16%. What did I do wrong?
    Last edited: Oct 26, 2003
  2. jcsd
  3. Nov 6, 2003 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I'm not sure what you're calculating. Think of it this way:

    (horizontal component of F)x (horizontal displacement) = KE(f)-KE(i)

    The initial KE(i) is, of course, the same. The only difference in the two cases is the horizontal component of F. Try again.

    Hint: No need for any formulas with m or v !
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