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Question concerning work and horizontal force.

  1. Mar 9, 2010 #1
    1. The problem statement, all variables and given/known data

    If it takes a horizontal force of 300N to push a stalled automobile along a level road at a constant speed, how much qwork must you do to push this automobile a distance of 5.0m?



    2. Relevant equations

    W = F x [tex]\Delta[/tex]x


    3. The attempt at a solution

    I used the above formula W = (300N)(5.0m) = 1.5 x 103 J, however, in checking my work with the solutions manual, they used the formula W=FsCOS[tex]\theta[/tex]. Both formulas come upi with the same answer.

    I do not understand why the authors chose to use the W=FsCOS[tex]\theta[/tex] formula because I thought that this formula was only used when "the motion of the particle and the force are not along the same line" (from the book), such as "along some arbitrary curved path."

    Is this a case where W = F x [tex]\Delta[/tex]x can only be used in cases where the motion of the particle and the force are along the same line but W=FsCOS[tex]\theta[/tex] can be used in either case?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 9, 2010 #2
    Yes, you are correct. The most general formula for work done on a body by a force is "the product of the force and the displacement of the body in the direction of the force"; in other words, the dot product of force and displacement, which can be represented as [tex]W = F\,s\, cos \theta[/tex] (strictly speaking, [tex]|F||s|cos\theta[/tex])

    In the case where the direction of the force is in the same direction as the displacement, we realise that [tex]\theta[/tex] is in fact 0. Hence, we obtain
    [tex]W = F\,s\,cos 0 = F\,s[/tex]
     
  4. Mar 9, 2010 #3

    Thank you for the explanation. You explained it much better than my text book did.
     
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