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Clarification on force on frictional surface constant speed

  1. Mar 18, 2015 #1
    for the following 2 examples from:
    http://www.physicsclassroom.com/class/energy/Lesson-1/Calculating-the-Amount-of-Work-Done-by-Forces

    A 10-N force is applied to push a block across a frictional surface at constant speed for a displacement of 5.0 m to the right.
    u5l1a15.gif
    Would it be more accurate to say that a 10-N force is applied to 'Maintain' (not to push or pull) a block across a frictional surface at constant speed for a displacement of 5.0m to the right?

    because the 10-N force is acting upon the frictional force canceling it out in order for the block to continue moving at its original velocity. the 10-N force isn't acting on the block to push it when it is canceled out by friction. the block is in equilibrium with constant velocity.

    also

    An approximately 2-kg object is pulled upward at constant speed by a 20-N force for a vertical displacement of 5 m.

    u5l1a17.gif
    Would it be more accurate to say that an approximately 2-kg object is 'moving' upward at a constant speed 'maintained' by a 20-N force for a vertical displacement of 5 m?

    because again, the tension force is acting not on the object but on the force of gravity. (any excess would be acting on the object). if a force is canceled out, it can't act on the object!!

    thanks.
     
  2. jcsd
  3. Mar 19, 2015 #2

    PhanthomJay

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    Your thinking is not correct. Forces act on objects, not on other forces. In your second example, the 20 N applied force acts upward on the object, and the 20 N weight force (gravity force) acts downward on the object. Together, the net force acting on the object is 0, so the object, once set in motion with a nudge, moves at constant speed , per Newton 1. Or in your first example, both the applied 10 N force and the opposite 10 N friction force act on the object. If you were the 'object', you would certainly feel an internal force acting on you, even though the net external force acting on you is 0.
     
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