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Work done on a wagon

  1. Dec 26, 2007 #1
    [SOLVED] Work done on a wagon

    1. The problem statement, all variables and given/known data
    A child and the wagon he/she is riding in has a combined mass of 50 kg, and the adult pulling the wagon does 2.2 x 10^3 J of work pulling the two 60 m at a constant speed. The coefficient of friction for the surfaces in contact is 0.26.
    a) draw an FBD for the wagon
    b) determine the magnitude of force applied by the parent.
    c) determine the angle at which the parent is applying the force

    2. Relevant equations
    Well the equations I am using seem to indicate that this is a nonsense question, but here goes:
    W=Fd (work equals force times change in displacement)
    F(kinetic)=F(normal)mu(K) (force of kinetic friction equals normal force times coefficient of kinetic friction)

    3. The attempt at a solution
    Since the amount of work is given, I divided it by the displacement and got 36.67 N, which should be the force applied by the adult in the direction of motion.

    However, using the formula for kinetic friction I get 127.4 N opposing the motion. If the wagon is moving at constant speed, shouldn't these two be equal? I don't really know where I am going wrong, since from what I understand "work" only applies in the direction of motion, and the friction force only applies in opposition to the direction of motion...needless to say this problem is confusing me, any help would be appreciated.
  2. jcsd
  3. Dec 26, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The trick here is that the force is applied at some angle, which changes the normal force. Draw yourself a FBD showing this force acting at some angle. Then make use of the two facts: (1) The work is given, and (2) The velocity is constant. Use them to solve for the force and the angle.
  4. Dec 26, 2007 #3
    ahhhh thats where you didnt read part c.

    Part c asks for the angle which the father is pulling, that means that the parent is applying a force that has a vertical component too, thus lowering the reaction force on the wagon. That means that you have to come up with an equation in part c.the kinetic friction experienced is also dependent on this vertical force that the parent is exerting on the wagon. You ahve to do a subtraction between the weight and the force.
  5. Dec 26, 2007 #4
    I think I've got it now.

    Friction force equals applied horizontal force which can be found using equation for work, and also equals given coefficient of friction x the normal force.

    Applied vertical force = Gravity force - Normal force

    Use pythagorean theorem to find applied force.

    For part c) angle = tan^-1 F(applied vertical)/F(applied horizontal)

    my answer is 351 N at 84 degrees to the horizontal
  6. Dec 26, 2007 #5

    Doc Al

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    Staff: Mentor

    Looks good to me.
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