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Work and Kinetic Energy of a Sled

  1. Jan 12, 2016 #1
    1. The problem statement, all variables and given/known data
    A man pulls a 15-kg sled containing a 5.0-kg dog along a straight path on a horizontal surface. He exerts a force of 55 N on the sled at an angle of 20.° above the horizontal, as shown in the figure above. The coefficient of friction between the sled and the surface is 0.22.

    a. Create a free body diagram for the forces on the sled

    b. Calculate the normal force acting on the sled

    c. Calculate the work done by the pulling force if the sled is moved 7m

    d. At some later time, the dog rolls off the side fo the sled. The man continues to pull with the same force. Sketch a graph of KE versus time t for the sled. Include both the sled's travel with and without the dog on the sled. Clearly indicate with the symbol tr the time at which the dog rolls off.


    2. Relevant equations
    W=FL
    W=cos@Fl
    Fn=mg-Fpullsin@
    Ff=u(mg-Fpullsin@)


    3. The attempt at a solution
    My thought is that the 55 N force is acting only on the 15kg of the sled, not the 5kg mass or a 20kg system. Therefore, the free motion diagram will have a 147 N (15*9.8) force of gravity acting on the sled, the 55 N force acting on a 20 degree angle, the normal force of the sled (147-Fpullsin@=147-19=128 N), and the force of friction, which does include the 5kg dog, which would be 0.22(20kg*9.8-55sin20)=38.6 N.

    The normal force on the sled would be 15kg*9.8-55sin20=128 N

    The work done by the pulling force on the sled is equal to the horizontal component of that force times the distance: (55cos20*7.0m)=361.8 N

    I am not sure how specific I need to make the graph, but it would appear to me that the Net force acting on the sled in the horizontal direction would increase from 13 N to 24 N before and after the dog roles off, respectively, because the horizontal component of force remains constant at 52 N but the resisting force of friction decreases from 39 N to 28 N. Therefore, the acceleration of the sled would increase from (13.1=15a) a=0.87m/s^2 to (23.54=15a) a=1.57m/s^2, so its kinetic energy, holding mass constant at 15kg, would be increasing by an exponential amount both before and after the dog roles off, but would increase even faster after this event because KE=mv^2/2, and if a is positive, v must be increasing linearly, which would translate to an exponential increase in kinetic energy through this equation.

    Is this right, or am I completely off base in my thinking?
    Can the graph of kinetic energy as a function of L be calculated exactly, or should I create a more general sketch?



     
  2. jcsd
  3. Jan 12, 2016 #2
    1. the force is acting on the sled with the dog on it. (you should allways think of the sled-dog as one body unless stated otherwise, like in the last question; with your reasoning pulling the sled with 52N force with or without the dog would produce the same acceleration, that does not make sense)
    2. the normal force on the sled with the dog on it must be considered with mass of the sled + mass of the dog

    since the problem says sketch id say it doesnt have to be very specific. the way you are thinkign about for the graph is correct i think. the level of precision... i dont know what to tell you

    dont forget to recalculate previous acceleration with the dogs mass on it, acceleration increases because there is less friction, bt also because there is less mass on the sled

    And mass would not be constant for Ke. it would also decrease
     
  4. Jan 12, 2016 #3

    haruspex

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    To offer a slightly different view from WrongMan's, it's ok to say the pull acts only on the sled as long as you consider all forces on the sled. It may be that the sled is accelerating. If so, there must be a frictional force accelerating the dog, and therefore an equal and opposite frictional force acting on the sled.
     
  5. Jan 12, 2016 #4
    Thank you so much everyone for your replies.

    I was not given any information regarding the friction between the dog and the sled, so I think this means that I am supposed to treat them as a single system.
     
  6. Jan 12, 2016 #5

    haruspex

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    Either way, you have to assume there is enough friction to keep the dog on the sled, until it got bored and rolled off. That is enough information.
     
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