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Sliding Block - True or False?

  1. Nov 1, 2013 #1
    1. The problem statement, all variables and given/known data

    It's in the picture attached...

    2. Relevant equations


    3. The attempt at a solution

    I picked options 2, 5 and 7 to be true but 2 turned out to be false. I can't figure out why it's wrong and an explanation would be appreciated. I mean it makes sense to me because work done by gravity is F*s which is (m*g)*h which is the same as the formula for gravitational potential energy, U = mgh

    Attached Files:

  2. jcsd
  3. Nov 1, 2013 #2
    i assume that you were using W=mgh and U=mgh? this is only correct when it is vertical, but since there is an angle θ, you need to use that other Work equation (W=Fdcos θ) which means that potential energy is not the same as Work. again, that would only be true if it were vertical such as a raindrop.
  4. Nov 1, 2013 #3
    Got it. Thanks
  5. Nov 1, 2013 #4
    Another Question !

    I was thinking of applying this logic to a pendulum situation.

    Let's say the pendulum is in motion and is at a certain moment in time at an angle of 45° to the vertical. I know that the work done by the tension is 0 as it's tangential to the direction of motion. What about the work done by gravity until the point where the pendulum is vertical?

    Is it mgh or mgh*sin θ? I think it's just mgh because the equation for work done by gravity was given to me as mg(L-L*cos θ) where L = Length of string at that moment in time and it follows that (L-L*cos θ) is = h.

    If it indeed is mgh, then it looks like we have assumed that the force of gravity only does work in the downwards direction. As tension does no work, what force does work in the direction of motion??
  6. Nov 1, 2013 #5

    Doc Al

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    Staff: Mentor

    Pay careful attention to the wording of option 2. When the block slides down, is the work done by gravity positive or negative? Is the change in potential energy positive or negative?
  7. Nov 1, 2013 #6

    Doc Al

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    Staff: Mentor

    Note that h = d cosθ, so the two expressions for work are equivalent.
  8. Nov 1, 2013 #7
    Oh. Does this mean that Eagle's answer was wrong and that the reason answer 2 is false is not because the formula of work is not mgh as Eagle argued but because change in potential energy is equal to - (work done by gravity) and not equal to work done by gravity itself?
  9. Nov 1, 2013 #8

    Doc Al

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    Staff: Mentor

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