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Understanding why change in Kinetic Energy is 0

  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known data

    Find the minimum force ##\displaystyle F## required to move ##\displaystyle M_2##.


    Details and Assumptions:
    ##\displaystyle M_1 = 3Kg##
    ##\displaystyle M_2 = 5Kg##
    ##\displaystyle \mu_1 = 0.4##
    ##\displaystyle \mu_2 = 0.6##
    ##\displaystyle g = 9.8m/s^2##


    2. Relevant equations


    3. The attempt at a solution



    The solution suggested is as follows:

    Let ##\displaystyle M_1## be displaced by a small distance ##\displaystyle x##

    Using the Work energy theorem,
    $$\displaystyle Fx - \mu_1M_1gx - \frac{1}{2}Kx^2 = \Delta K.E.$$

    For the situation that ##\displaystyle M_2## just moves,
    ##\displaystyle \Delta K.E. _{M_1}= 0##

    However, I cannot understand why the change in Kinetic Energy of M1 should be 0. How do we know that when M2 is just about to move, M1 will be at rest?

    I've been troubled by this doubt for ages and would be really grateful if somebody would kindly explain the reason to me in detail. Many many thanks in advance!
     

    Attached Files:

  2. jcsd
  3. Sep 21, 2015 #2

    jbriggs444

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    It is baffling to me why someone would invoke the work energy theorem here. It appears that the approach is to use spring energy computed by integrating Hooke's law (F=kx) over distance to obtain energy (E=1/2kx^2) and then to differentiate that (hence the "small distance x") to obtain force.

    Why not do it the easy way and simply compute force from Hooke's law directly?

    Edit: Never mind me. I see I've misunderstood the scenario. The optimal scheme involves getting M1 up to a decent speed.
     
  4. Sep 22, 2015 #3
    Sir, please could you explain why ##\Delta K=0##?
     
  5. Sep 22, 2015 #4

    haruspex

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    To move M2 you require a certain force exerted by the spring. That corresponds to a certain compression of the spring. Once that compression is reached, M2 will move - you do not need any further compression. So you only need to move M1 far enough to achieve that compression.
     
  6. Sep 22, 2015 #5
    Thanks very much Sir. Sir. please could you help me with another question too?
    https://www.physicsforums.com/threa...ing-on-an-inclined-plane.833831/#post-5235629
     
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