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Work and Zero Work

  1. Mar 18, 2012 #1
    Hi, I am taking a highschool physics course and I just need a little help grasping the concept of zero work and whether or not it applies to this situation... The question reads:

    A child on a sled (having a combined mass of 47.0kg) is pulled by a force directed along a rope that make a 45° angle with the horizontal axis. The force exerted on the rope is 100.0N. The force of friction acting on the sled is 30.0 N. If the child is pulled a distance of 10.0 m along a level field, determine the total work done on the child and on the sled.

    So I worked this out using W = F(cos θ)• ∆d
    I calculated the work from the pull and then the work from friction and added the values to get 407 J which I'm pretty confident is right..

    Where I need help is in deciding whether or not this value is the work done on the child and the sled..
    is this the work done on the sled. And the work done on the child is zero. I'm at this dilemma because my text says that 'whenever a force is exerted perpendicular to the direction of displacement, it does not contribute to forward motion.'
    So I'm thinking that the child is just kind of along for the ride and the sled is experiencing work. I think this because the child is only experiencing ForceNormal and ForceGravity.. but then again isn't the child experiencing some kind of ForceFriction from the sled and does the fact that the child's weight is definitely affecting the answer something to consider? Please help!
  2. jcsd
  3. Mar 18, 2012 #2


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    What you've worked out is the work done on the combined system of child+sled. And that's perfectly legit. There is no reason why you cannot consider the child and the sled to be a single object having their combined mass, since the child and sled are rigidly attached (more or less). When one moves, so does the other.

    What keeps the child and the sled attached to each other? Friction, of course. Friction between the child and the seat. Without this friction, the sled would slide out from underneath the child when pulled on. In fact, since you know that the acceleration of the child is the same as the acceleration of the overall system, you can work out the net force on the child (due to this friction) and hence the work done on the child alone.
  4. Mar 18, 2012 #3


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    I think that's right.

    The sled and the child are accelerating, because of work being done on the sled. Assuming the child is not moving relative to the sled, the child and sled both have the same change in velocity and they both have mass. So the kinetic energy of both is increasing.

    Each kg of mass of the child+sled has the same increase in velocity, so each kg of mass has the same increase of KE. You know the total increase in KE (= the work done on the system) so you can split the total into the work done on the child and the work done on the sled.

    If you want to work this out using forces, the sled is applying a force to the child to accelerate it (i.e. the friction force between the child and the sled), and the child is applying an equal and opposite force to the sled. But you can answer the question using energy without finding the magnitude of that force, and that is one reason why "energy methods" are useful in mechanics.
  5. Mar 18, 2012 #4
    Alright, so far I'm following..
    Couple things though.. if I drew a FBD of the child.. in regard to just the horizontal forces..
    would the child have both Forceofsled and Forcefriction pulling it forward.. is that right?

    also im confused as far as how to split the forces between the sled and the child..

    so far I'm thinking:
    I feel like the question doesn't give me enough information. Dont I need to know either the coefficient of static friction or forcenormal? and as far as I can tell I need to know the mass of the child without the sled in order to calculate the child's Forcenormal.
  6. Mar 18, 2012 #5


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    What is "forceofsled"?

    Thanks to friction between the sled and the child, the child is accelerating along with the sled. That's it. That's the only horizontal force acting on the child.

    Enough information to do what? You've already solved the problem, haven't you? You've found the work done on the combined mass of the child+sled.

    In order to find the work done on the child alone, you'd need to know the mass of the child, it's true. If you did know that, then you would be able to compute the work done on the child without computing any of the forces on the child at all, as AlephZero has already pointed out. You can just use energy methods.
  7. Mar 18, 2012 #6
    sweet baby jesus.. I understand

    Thanks so much to the both of you!
  8. Mar 18, 2012 #7
    one more thing.. part b) asks..determine the childs final speed at the end of 10.0m
    does this look right?

    W = ∆Ek
    W = Ekf - Eki
    W = (1/2)mvf^2 – 0 J
    407 J = (1/2)(47.0kg) vf^2
    vf = √(814/47.0)
    vf = 4.16 m/s
  9. Mar 18, 2012 #8


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    That looks fine to me.
  10. Mar 18, 2012 #9
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