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Homework Help: Work problem. Please help.

  1. Sep 30, 2007 #1
    A factory worker pushes a crate of mass 28.5 kg a distance of 4.70 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.25.

    What magnitude of force must the worker apply?

    How much work is done on the crate by this force?

    How much work is done on the crate by friction?

    I know I have to use the equation W=[tex]\vec{F}[/tex]*[tex]\Delta[/tex][tex]\vec{r}[/tex] for work. Then work gravity = mghcos0 and i got wgravity=4.7*28.5*9.8=1312.7 and i know thats wrong. Can someone please help me with this?

  2. jcsd
  3. Sep 30, 2007 #2


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    Remember that in the dot product formula, [tex]\theta[/tex] is the angle between the direction of the displacement (the direction the object moves) and the direction along which the force is applied. Which way does the gravity (weight) force point relative to the direction the crate is being moved?

    It will be helpful for you to make a free-body diagram for the crate, showing the forces acting on it, since you will need to see the directions they point to answer the rest of these questions.
  4. Sep 30, 2007 #3
    That's right.

    What's "work gravity"? The crate is being pushed on a level floor. Have you figured out the force the worker applies?

    I suggest drawing a free-body diagram.

    Edit: Ah, dynamicsolo beat me to it. :)
  5. Sep 30, 2007 #4
    Ok. I drew a free body diagram. The Fn and g cancel out leaving the F right? So I know that mk=.25. So is it just F=.25*28.5?
  6. Sep 30, 2007 #5
    The normal and gravitational forces cancel, but there are two forces acting along the horizontal direction. The crate is moving at constant velocity. What does that tell you about the net force acting along the horizontal direction?

    No. Force = dimensionless constant*mass - that is dimensionally incorrect.
  7. Sep 30, 2007 #6
    i got it to be 70N. I did 28.5*.25*9.8.
  8. Sep 30, 2007 #7
    W= 70 * 4.7 which is 330 J
  9. Sep 30, 2007 #8
    how would i find how much work is done on the crate by friciton? Would it just be W=70*.25?
  10. Sep 30, 2007 #9
    That's the magnitude of the force of friction, and since the crate is moving at a constant velocity, it should also equal the mag. of the force applied by the worker, which leads to the work done by him over a distance of 4.7m to be...
  11. Sep 30, 2007 #10
    Nope. You just gave the definition of work in your first post. Also refer to dynamicsolo's post.
  12. Sep 30, 2007 #11
    OK. So the work of friciton would be the same except in the opposite direction which is -330 J. So the work done by the normal force would be 70J too?
  13. Sep 30, 2007 #12
    See the definition of work you gave in first post, whats the angle between normal force and disp.
  14. Sep 30, 2007 #13

    The net work done on an object is equal to change in the kinetic energy of the object. As you can see, the crate has zero velocity, and therefore zero KE. So the net work done on it must also be zero, which means work done by worker and that done by friction should cancel out to zero. (i.e. 330 + (-330))

    Again, use the definition of work. What is the direction of the normal force and what is the direction of displacement?
  15. Sep 30, 2007 #14
    The direction of the normal force is acting downward on the box and the displacement is to the right of the box.

    So Fk=mk*Fn

    the normal force = 28.5*9.8 / .25?
  16. Sep 30, 2007 #15
    Fn & displacement are vectors. If angle between two vectors is 90, their dot product is zero. And Normal force on crate by floor acts upward, not downward(because if it is downward then the block will have net downward force and it will fall down).
  17. Sep 30, 2007 #16
    Ohhh. Ok. So to find the gravitational force, it would be the same thing? their dot product would be zero?
  18. Sep 30, 2007 #17
    And that means that the total work done on the crate would be zero too?
  19. Sep 30, 2007 #18


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    Hurrah! Yes, since the crate is accelerating neither vertically nor horizontally as it travels on a level surface, there is no net work done on it. If you've gotten to the work-kinetic energy and work-potential energy theorems, you'll see that this is just as we expected. (If you haven't gotten to them in your course, you're probably about to.)

    You now want to go back and make sure you have answered all the questions in the original problem.
    Last edited: Sep 30, 2007
  20. Sep 30, 2007 #19
    Thank you guys for all your help! and bearing with me threw my problem!! :)
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