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Work, forces on inclined plane

  1. Oct 5, 2013 #1

    W = fd

    The answer is 8. The solution manual says work is the same for either scenario, d = 4 or d = 0.5

    From this assumption I can see that by increasing distance from 0.5 to 4, you reduce the force necessary by a factor of 8.

    However what I cannot comprehend is: where does the assumption that work is the same for both scenarios come from? Since you have the added horizontal component to consider when pushing it up the ramp, isnt that extra work? Or....Have I labeled the triangle incorrectly? Should the 4 be labeled on the other leg rather than the hypotenuse?

    Sorry if this is a silly question but I havent taken intro physics in a couple years and am currently studying for the mcat. It seems not all of the principles of physics have hstayed with me lol. Thank u in advance pf!

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Oct 5, 2013
  2. jcsd
  3. Oct 5, 2013 #2
    You can analyze this problem in this way

    The min force required to lift an object of mass m vertically up from bottom to top of the step is F1= mg

    The min force required to move an object of mass m up along the ramp from bottom to top of the step is F2= mgsinθ

    From this you can deduce F2=F1/8 .

    Does this help ?
  4. Oct 5, 2013 #3
    Yasssss! Thank you so much!!!!
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