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Basic physics discussion

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data

    True or False: To raise a 2.3kg mass from its resting place on a table, more work is done by lifting it diagonally than lifting it straight up.

    2. Relevant equations

    w=fd


    3. The attempt at a solution

    True: I argued that although the same amount of force is being applied vertically, some force needs to initially be applied in the horizontal component, thus making the overall work greater.

    Correct answer is false
     
  2. jcsd
  3. Nov 20, 2008 #2

    LowlyPion

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    Homework Helper

    Gravity is a conservative field.

    Work goes to potential energy which is given merely by the m*g*Δh without regard to the path it took to get there.
     
  4. Nov 20, 2008 #3
    Your argument is actually a pretty good one. Some work is indeed done by the force applied initially. But that is not the entire story - what happens after you lift the mass?
     
  5. Nov 20, 2008 #4
    The question didn't even specify it was a frictionless system, so there would be air resistance as well.

    @Lowly Pion: I'm not sure I understand. Although work is still being done and converted to gravitational potential energy, wouldn't an additional force have to be applied to start horizontal translation. I guess you could say that it would be an infinitely small force over an infinitely small interval of time, but it is still a force nevertheless.

    @Naresh: What does happen after you lift the mass? In both cases they would have the same gravitational potential energy, just in one case slightly more work was done to achieve it.
     
  6. Nov 20, 2008 #5
    If there is air resistance, you obviously do some additional work since the field is not conservative - but let us ignore friction for now.

    So you did some additional work and gave the object some horizontal velocity. What happens to that horizontal velocity? The additional work you did was not to achieve the change in gravitational potential energy..
     
  7. Nov 20, 2008 #6
    I agree.

    Since the force vector is always pointing donwards, due to acceleration produced by gravity, then the work that is being done in both cases is the same.
     
    Last edited: Nov 20, 2008
  8. Nov 20, 2008 #7
    I'm afraid I don't understand that argument.The horizontal velocity will remain constant, but I don't see that being relevent since a force was still required to start it. While the additional work done didn't achieve any gravitational potential energy, additional work was done.
     
  9. Nov 20, 2008 #8
    It doesn't matter. Look at the equation:

    W=F*d
    If you lift it a certain distance straight up, the force required will be greater than if you had to lift it diagnally. However, if you lift it diagnally, less force is required, HOWEVER, you cover a greater distance, which compensates for the smaller force.
     
  10. Nov 20, 2008 #9
    So if I lift a brick a foot off the ground diagonally over a 1000m distance, it would take the same amount of work to lift it straight up?

    I have done the same work vertically, but I have also done fd horizontally. (Sorry that was poorly worded)
     
  11. Nov 20, 2008 #10
    Just looking at the horizontal component of the displacement, force, velocity etc.:

    Initially you apply a small force to get the object moving in the horizontal direction. This does work, and in the absence of friction, imparts some velocity to the object. The work, therefore, has gone into increasing the kinetic energy of the object.

    When the object eventually comes to a rest, it has no horizontal velocity. Its kinetic energy has been removed, therefore some negative work has been done on it. This means you applied a force in the reverse direction. Again, in the absence of friction, the negative work done to stop the object is equal in magnitude to the work done to get the object moving horizontally. Therefore, if you move a brick a km horizontally, you don't do any net work. (In your equation W = F.d, F is not constant, it changes sign). The only work done goes towards increasing the potential energy of the object.

    This may be hard to imagine because there is always friction in the real world, so one is always doing some work to overcome energy dissipation. But in an "ideal" world, there is no work done when something is moved horizontally.
     
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