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Work-Kinetic Energy Theorem

  1. Dec 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A 2100 kg car accelerates from rest at the top of a driveway that is sloped at an angle of 20.0º with the horizontal. An average friction force of 4000 impedes the car's motion so that the car's speed at the bottom of the driveway is 3.8 m/s. What is the length of the driveway?


    2. Relevant equations

    Wnet=Change in Kinetic Energy
    KE=.5mv^2
    Work=FDcos(theta)

    3. The attempt at a solution
    FDcos(theta)=.5mv^2(final)-.5mv^2(initial)
    4000(2100)(9.8)dcos(20)=.5(2100)(3.8)^2
    d= ((3.8)^2/(4000(9.8)cos(20)))
    d= Some tiny decimal.
    The book answer is 5.1m
    Where did i go wrong, and how can i fix it?
     
  2. jcsd
  3. Dec 7, 2009 #2
    The net force is not 4000N. It is the force of gravity pulling the car down the hill minus the force of friction. Recalculate with this value of F and you will get the right answer
     
  4. Dec 7, 2009 #3
    I'm still not getting it.
     
  5. Dec 7, 2009 #4
    According to Newton's 2nd law of motion, the net force on an object is given by [tex]\sum F=ma[/tex].
    In your equation [tex]W=Fdcos\theta[/tex], F is the same as [tex]\sum F[/tex] in the equation for Newton's 2nd law. In the case of the car on the incline, [tex]\sum F=F_{gravity}+F_{friction}[/tex].

    You have everything right accept for the force. In your case, you have [tex]\sum F=F_{friction}[/tex].

    All you have to do is figure out what the force of gravity is. Think back to a block sitting on an incline and you had to figure out the forces in the x- and y-directions (Im sure you had to do this at some point). When you calculated the gravitational force in the x- and y- directions you had to take into account the angle of the incline. Try this and you should get the right answer. If not, I will check back in a little while and help you out some more.
     
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