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Homework Help: Force of friction

  1. Jul 28, 2011 #1
    1. The problem statement, all variables and given/known data

    A 25 kg block is pulled up a ramp 20 meters long and 3 meters high by a constant force of 120 N. If the box starts from rest and has a speed of 2 m/s at the top, what is the force of friction between the box and the ramp?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jul 28, 2011 #2


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    Use the work-energy theorem:
    The work done on a body by the net force is equal to the change in its kinetic energy.​
  4. Jul 28, 2011 #3
    Or, the work done by all the non-conservative forces is (negative) equal to the change in the total mechanical energy (kinetic + potential energy). Then, work done by a constant force is the product of the force and the displacement along the line in which the force is acting. If the angle is obtuse (opposite directions), then this work is negative.
  5. Jul 28, 2011 #4
    please in formula? I cant understand the question coz i hadn't got any sleep yet sorry
  6. Jul 28, 2011 #5
    Use the work-energy theorem : The work done by all forces on an object = change in it's kinetic energy.
    (1) You've got the final and initial velocities, so you can get the change in kinetic energy.
    (2) All the forces here are constant, so you can directly use F.s
    (3)The forces are the applied force, friction, weight and normal reaction.
    (4) Choose the coordinate system, preferrably, with the an axis on the slope. You've also got the angle of the slope.
    (5) You know the displacement. So, find out the work done by each force by using F.s = F.s.cos(angle between the F and s), and equate it to the change in kinetic energy.
    You have one unknown, the force of friction, so one equation will take you through!!
  7. Jul 28, 2011 #6

    thank you very much for helping me, now i can sleep.
  8. Jul 28, 2011 #7
    Good night. What's the answer by the way?
  9. Jul 28, 2011 #8
    hehehe im just answering it, but i think ill be over in no time . ill just post the answer right after.

  10. Jul 28, 2011 #9
    im skeptical but i found out that its 2343.779092

    seems big for a friction force? . . . is it?
  11. Jul 28, 2011 #10
    This is not an acceptable result for a force.
  12. Jul 28, 2011 #11
    yeah you're right urrrghh ill just re-compute it when i wake up this morning, the dawn is getting over my brains.

    may be a few hours or sleep would wake my brain up, well thanks for noticing my mistake. good day!!
  13. Jul 28, 2011 #12


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    Alternative 1.
    The body will gain a certain amount of Potential energy - it is higher in elevation at the top of the slope than at the bottom.
    The body has gained a certain amount of Kinetic energy - it is moving faster when it reaches the top.
    In the absence of friction, what force would you need to achieve that? - lets say it was 85 N.
    The friction Force must be the difference between the actual applied force and the required force - so if 85 was correct - friction would be 35N.

    note: if when you work it out, 85n is the right interim answer, you may toast my guess - I just made up a number that was less than 120!!

    Alternative two:

    Energy is added to the block when you apply the 120N Force along the full length of the slope.

    Energy is taken from the block by the friction force, which also acts along the full length of the slope.

    The gain in energy the block has is partly because it is higher up [mgh] and partly because it is now moving at a speed [1/2mv^2]

    you can easily calcuate the Work done by the applied force, the Kinetic Enegy gained and the potential energy gained. From that you can calculate how much work is done by friction - and therefore the size of the Friction Force.

  14. Jul 28, 2011 #13


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    Moderator's note:

    rjulius: please show your work.

    Others: please wait for rjulius to show his work before offering further help.
  15. Aug 1, 2011 #14
    Did you get enough sleep already? You never posted the correct solution.
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