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Two questions concerning mechanics

  1. Oct 29, 2005 #1
    I have a question about certain conditions in Newtonian mechanics...
    Let's say you had an incline, with a pulley at the edge of the top, frictionless, let's keep the string & pulley masses negligible, and there's a light string connecting an object hanging, and an object on the incline...
    I know that all of our assumptions that tension is equal is true, but how is the tension equal throughought? Isn't it accelerating a different value on both sides? I know I'm probably asking a stupid question, but if the two masses are different (one is 3kg, one is 9kg, for instance) the acceleration on one due to the tension will be more than the other, wouldn't it? Or would it not?
    Can someone explain this to me with proof behind the explanation? (Again, apologize if this is a stupid question)

    Also, let's say a projectile was thrown with some initial velocity. At its highest point, when we consider its potential energy, (Let's say it has a constant horizontal velocity, i.e. it was given that initial velocity at an angle in the first place) do we consider it by considering both the horizontal potential and kinetic energy as well as the vertical potential and kinetic energy, (by using vector addition and the like) or do we consider them to be two different things and ignore the horizontal direction?
    Can someone explain this with proof and perhaps an example?

    Thanks a lot for any responses.
  2. jcsd
  3. Oct 30, 2005 #2

    Doc Al

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    Staff: Mentor

    Realize that as long as the string does not stretch, both masses are constrained to move together with the same speed and acceleration. What determines the acceleration of each mass is not merely the tension (which is the same throughout since the string is massless) but the net force on each mass, which includes gravity.
    Potential energy depends only on the vertical position (height), so a "horizontal" PE makes no sense. Near the earth the formula for gravitational PE is mgy, where y is measured from some arbitrary reference point.

    KE energy depends on speed, not direction. But since [itex]v^2 = v_x^2 + v_y^2[/itex], it's OK to think of the KE as associated with the vertical and horizontal components of velocity separately, if it helps you. At the top of the projectile's motion, the only speed is horizontal so [itex]{KE} = 1/2 m v^2 = 1/2 m v_x^2[/itex].

    Both PE and KE are scalars, not vectors, so you just add them like ordinary numbers. Using vector addition makes no sense, since energy does not have a direction.
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