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Work done incline block system

  1. Sep 10, 2015 #1
    1. The problem statement, all variables and given/known data
    A block of mass m is released from rest on a smooth inclined plane (wedge). The wedge itself is resting on a smooth horizontal surface. The block undergoes a vertical displacement of h. Assume that the height of the incline is h and the length of the incline is h cosec θ

    Work done by normal reaction force on the block exerted by the wedge is--
    (A) Positive
    (B) Negative
    (C) Zero
    (D) Less than mgh in magnitude

    Work done by normal reaction force exerted by block on the wedge is--
    (A) Positive
    (B) Negative
    (C) Zero
    (D) Less than mgh in magnitude


    More than one choices may be correct.


    2. Relevant equations


    3. The attempt at a solution
    I'm so confused over this basic problem. Firstly, I know that the wedge will move a little because it is a smooth surface. Answer to the second question should be positive, because there will be a component of the normal in the opposite direction of the motion of the block. So the wedge will move in the direction of the horizontal component of the normal exerted by the block. But I don't know why the work will be less than mgh.

    For the first question, I'm very confused.
     
  2. jcsd
  3. Sep 10, 2015 #2

    andrewkirk

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    What formula have you been taught about the relationship between force, motion and work?

    Does a force do work on an object when the direction of motion (of the object upon which the force acts) is perpendicular to the direction of the force?

    Is that the case in one or both of the questions? Which one?
     
  4. Sep 11, 2015 #3
    That's not the case in either of the two questions, because the wedge itself moves a certain distance. The work done is not zero in any case. For the second one, it's positive. For first, I'm not sure because that would depend on how much the wedge moved (I think).
     
  5. Sep 11, 2015 #4

    andrewkirk

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    Fair enough. Perhaps looking at the immediate forces is not the best approach.

    In any case it sounds like you've satisfied yourself as to the answer on the second one, and concluded that the wedge starts moving in the opposite direction to which it points, but are not sure whether the block is moving in the same direction, in the opposite direction, or not at all, relative to the floor. Could conservation of momentum help with that?
     
  6. Sep 11, 2015 #5
    Consider mgh your "bank account" of available energy. How is that energy "spent"?
     
  7. Sep 11, 2015 #6
    Yeah I conserved momentum and I got the horizontal displacement of the wedge, which is less than hcotΘ. So overall the displacement of the block will be opposite to the horizontal component of the normal. Therefore, first is negative.
     
  8. Sep 11, 2015 #7
    Don't I need to know what the normal force is? And then take the dot product of the force and displacement? To find the normal force, would I have to consider pseudo forces due the the motion of the wedge?

    Do you mean by energy conservation? And if I thought about energy conservation, wouldn't mgh be used to bring the block down? But the block is coming down and moving horizontally.
     
  9. Sep 11, 2015 #8

    andrewkirk

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    Can you elaborate on how you concluded that? If the momentum of the wedge is leftwards (wlog) then won't conservation of momentum require the horizontal component of momentum of the block to be rightwards? Isn't that the same direction as the horizontal component of the normal force the wedge exerts on the block?
     
  10. Sep 11, 2015 #9
    Yeah it is but it's less overall. If you consider the initial x coordinate of the block to be 0 and the x coordinate of the COM of the wedge to be -c, then using the fact that COM won't change (and assuming m is mass of block and M mass of wedge and x is the horizontal displacement of the wedge),

    -Mc + 0 = M(-c+x) + m(-hcotΘ+x)
    x=(mhcotΘ)/m+M
    m/m+M is a fraction, so x will always be less than hcot Θ.
    This means that overall displacement would be hcotΘ-x in a direction opposite to the horizontal component of the normal.
     
  11. Sep 11, 2015 #10

    andrewkirk

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    I agree up to and including the second last line.
    What is the reasoning behind the step to the last line?
    For clarity assume the wedge is moving left, and the block is sliding down the wedge towards the right relative to the wedge, and the positive horizontal direction is rightwards.
     
  12. Sep 11, 2015 #11
    Okay I'm sorry I misconcluded, because overall the horizontal component does positive work because the displacement of the block is hcotθ - x in the positive x direction.

    But the vertical component of the normal does negative work, because the vertical component of the normal is oriented towards the positive y direction, but displacement of the block is in the negative y direction.

    But what is the overall work done? Is it positive or negative?
     
  13. Sep 12, 2015 #12
    Consider the limit where theta=90o. There is your answer.
     
  14. Sep 12, 2015 #13
    Negative but how can you say for every other angle?
     
  15. Sep 12, 2015 #14
    Hmmm...I see the problem. I would say a falling block is doing positive work, i.e., if it were hooked up to a machine, it could produce heat, say, and if free-falling would produce kinetic energy that is converted to heat when it hits the ground. In your case, I would agree with you that the block does positive work on the wedge to accelerate it and the wedge does negative work on the block to slow it from the velocity it would have achieved if the wedge were stationary.
     
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