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Work Energy Principle

  1. Aug 6, 2015 #1
    Hi guys, really confusing myself over grasping this principle... I think my main problem is understanding datums and how choosing the direction of a positive axis effects the result.
    I understand that the work energy principle states:


    Wnet = ½m(vf2+vi2)

    I have previously been taught that Wnet = losses - gains.
    Whilst this worked for my previous studies, its now proving that I need more of a clear understanding in order to approach the question correctly.


    1. The problem statement, all variables and given/known data
    it is given that:
    Ma = 30kg
    Mb = 20kg
    μk = 0.1

    we are asked to work out the speed of block A, given that B has moved up the plane 0.6m.

    (the poorly drawn image of the problem is below)

    5lLEctA.jpg

    please note, that the string on the pulley is light and inextensible, the lengths Sa and Sb I labelled since these vary.

    3. The attempt at a solution

    I have had countless of attempts... really im looking for an approach and a way of thinking that will not only allow me to solve this problem, but countless of others like it. Meanwhile Ill try solve it some more...

    My most recent attempt was to think about each object separately...

    - I first differentiated the 2 varying lengths of the rope to find a relationship between the velocities... I found that 2Va = -Vb (my question here is, since they are negated how do I go about treating the reference axis for both block A and block B, since they are two differently inclined slopes)

    - After which I also said for 2×ΔSa + ΔSb = 0... this then gives us a relationship of the change in distance... we know ΔSb = 0.6, therefore the other length is -0.3

    - now this is where I get a bit confused... As I stated, i recently considered each object separately... then my thought process was that for any force opposing motion, then this will give us a negative work... so formulated 2 separate energy equations for block A and B...

    - I thought that since block B is going up the incline, the tension force in the rope will be a positive work, and the weight and frictional forces will provide a negative work... then let this equal the final kinetic energy for B, I subbed in 2Va = -Vb however I ignored the signs, since I was just assuming a positive velocity up the slope (again bit confused here)...

    - I did the same for block B, however the tension work 2T*0.3 I took away from the net work, since it's opposing block A's motion... I added the work from the weight and deducted the work from friction.

    (another question is... forces opposing motion are meant to be doing negative work, this is why I added the work done by the weight on above... however gravitational potential energy is meant to decrease as you get lower in height, how does this fit in with my way of thinking with regards to the negative work?)

    - So after getting the two equations I added them together, this cancels out the work by the tension, since ones positive and the others negative. However after solving all of this, the final velocity that I calculate turns out to be wrong...

    Thanks to anyone who takes the time to read and answer my question(s)...

    My working:

    So I figured: [tex] 2S_{a} = -S_{b}, thus: 2V_{a} = -V_{b}. [/tex]

    for Block A:

    [tex] \mu (m_{a}g\cos{\theta_{a}})*S_{a} + 2T(S_{a}) = S_{a}\sin{\theta_{a}}*m_{a}*g - \frac{1}{2}*m_{a}*V_{a}^2 [/tex]

    for Block B:

    [tex] \mu (m_{b}g\cos{\theta_{b}})*S_{b} - T(S_{b}) = S_{b}\sin{\theta_{b}}*m_{b}*g - \frac{1}{2}*m_{b}*V_{b}^2 [/tex]

    however I sub in the previous relationship [tex] 2V_{a} = -V_{b}. [/tex]
    I ignore the sign however, due to me treating each system separately:

    [tex] \mu (m_{b}g\cos{\theta_{b}})*S_{b} - T(S_{b}) = S_{b}\sin{\theta_{b}}*m_{b}*g - \frac{1}{2}*m_{b}*(2V_{a})^2 [/tex]

    then I added the two equations together... giving me:

    [tex] \mu (m_{b}g\cos{\theta_{b}})*S_{b} + \mu (m_{a}g\cos{\theta_{a}})*S_{a}= S_{a}\sin{\theta_{a}}*m_{a}*g - S_{b}\sin{\theta_{b}}*m_{b}*g - \frac{1}{2}*m_{b}*(2V_{a})^2 - \frac{1}{2}*m_{a}*V_{a}^2 [/tex]

    then I finally solve for [tex] V_{a} [/tex] and get the wrong answer :(


    Where am I going wrong!?!??!
    Any general input on this topic would be really appreciated... its driving me nuts.
     
    Last edited: Aug 6, 2015
  2. jcsd
  3. Aug 6, 2015 #2
    I think that if you post your equations,finding the mistake would be easier.Meanwhile,what do you think of a kinematics approach to solve this problem?
     
  4. Aug 6, 2015 #3
    Okay, I'll write up my working now. I could approach it using kinematics, however it bugs me that I don't have a firm understanding on work energy, so would like to solve it this way :)
    Hope you can spot my errors when I update, thanks
     
  5. Aug 6, 2015 #4

    haruspex

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    I feel your main problem is that you need to state explicitly which directions are being taken as positive on each side and make sure you stick to that convention.
    '+'?
    Since the two velocities are in quite different directions, you clearly don't mean that as a vector equation. So whether they are connected by a plus or minus sign depends entirely on which directions you choose as positive. You could choose positive as up on both sides, or down positive both sides, or, since you can predict that B moves up, you might find it more natural to choose positive as up on that side and down on the A side. Whichever you choose, make sure you apply it consistently to all displacements, velocities and accelerations. It also helps to state it for other readers.

    g itself can be confusing. It is common to take up as positive. With that convention, it is logical to use 'g' for gravitational acceleration, but, when the time comes, to substitute g=-9.8 ms-2, or whatever. Most people seem to put -g for the acceleration, preferring the 'variable' g to have a positive value. Again, it's your choice, but make sure you are consistent.
     
    Last edited: Aug 7, 2015
  6. Aug 7, 2015 #5
    Thanks for your response haru, I feel it does answer my question... however I still am finding it hard to grasp the sign convention for everything...

    I have chosen up as positive for block B and down as positive for block A.
    This leads to a positive displacement for both block A and B right? so Sa = 0.3, Sb =0.6
    So the negative work for particle A would be the tension (2T) in the string, and the friction. For B it would be the weight and the friction...
    The velocity of block A with respect to B as said before is -2Va = Vb... However even if I take into account the sign, it makes no difference since its squared, so becomes positive.
    When solving then for A by combining the 2 equations and cancelling out the tension work, I still end up with the incorrect answer (I got velocity of A as 1.73m/s)

    I get on fine with near all of these work energy principle problems, however for some reason I seem to apply everything I know in this question and I still f'it up
    I think im just misunderstanding some key fundamentals, how do you go about defining the signs to use etc. within a problem...

    Thanks again :)

    oh and, that top equation was a typo should be a minus ;)
     
  7. Aug 7, 2015 #6

    haruspex

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    I believe you have a sign error in your block B energy equation in the OP. Check the ##\Delta PE## term.
     
  8. Aug 7, 2015 #7
    Hmm, so your saying I shouldn't deduct the work due to weight with respect to block B? I thought that since the weight is opposing block B's motion this would be a negative work?
    I added the work due to weight from block B just to see, and it still doesn't yield the correct answer.
     
  9. Aug 7, 2015 #8

    haruspex

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    The only source of work is the descent of A. All the rest are where that work goes. For B, T Sb = work done against friction by B + work done against gravity in raising B + work done in accelerating B.
    What answer do you now get? What is the answer supposed to be?
     
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