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The Cart's work

  1. Dec 5, 2009 #1
    A horizontal force of 200 N is applied to a 55 kg cart across a 10 m level surface. If the cart accelerates at 2 m/s2, then what is the work done by the force of friction as it acts to retard the motion of the cart?

    * -1100 J
    * -900 J
    * -800 J
    * -700 J




    I am confused with this problem and need help. I could use the help a.s.a.p., and I need ti finished by tomorrow night.
     
  2. jcsd
  3. Dec 5, 2009 #2

    LCKurtz

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    Hint: How much of the force is accelerating the cart?
     
  4. Dec 6, 2009 #3
    I would think all 200 N would be used to push and accelerate the cart.
     
  5. Dec 6, 2009 #4
    No. If there were no friction F=ma should obtain. Does 2m/s-s = F/m?

    So we need to expand our consideration of forces to include a frictional force, ie

    Fnet=(200-Ff). Work is defined as what mathematically? Ff is the force due to friction.
     
  6. Dec 6, 2009 #5
    I think it is defined as:

    W = F * [tex]\Delta[/tex]x
     
  7. Dec 6, 2009 #6
    Good. So you just need to compute the Ff. You have enough info, remember the force net = ma where a is the actual acceleration.
     
  8. Dec 6, 2009 #7
    I got F to be 110 N. So I take that and multiply it by the distance (10 m)?
     
  9. Dec 6, 2009 #8
    Not quite: a=2 mass=55 so ma=110=(200-Ff)
     
  10. Dec 6, 2009 #9
    So that leaves me with 90 N. (200 N - 110 N)
     
  11. Dec 6, 2009 #10
    Perfect, now multiply by distance. By the way since the force is acting in the direction opposite to that in which the mass is displaced, it will be negative. Your simplified eqn above omits the fact that it is a dot product, and direction determines work by a factor cos(THETA). In other words if you move perpendicular to the force, work is zero, if you move in direction parallel to force, work is positive and antiparallel, negative.
     
  12. Dec 6, 2009 #11
    Okay.

    Quick question: Since I have to show my work, how and where would I show that it becomes negative?
     
  13. Dec 6, 2009 #12
    It would appear in the dot product--I should switch to latex for this but in a hurry,

    W=Integral (F) cos(theta) dx where theta is 180 degrees. The other way is just to realize that since friction is a dissipative force (takes energy away from the system), it will always be negative. In other words whatever direction you drag the object, friction will be in opposition and dissipating energy. Also, had we considered the forces as both positive initially, 200=110+Ff which is appropriate, the negative sign would have been present. My bad on that.
     
  14. Dec 6, 2009 #13
    Okay, I see it now. Thanks. Now back to my Max. Net Force problem, and Speed of the Mass! (Thanks for all your help, by the way. Don't feel that I am taking advantage of it, or you. :smile:)
     
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