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Kinetic energy

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    The 4kg weight is released when the question begins


    2. Relevant equations
    kE = 1/2mv^2


    3. The attempt at a solution
    Had a go using SUVAT equations and came up with a velocity of around 22ms ready to put into the kE formula, then realised I hadn't done anything with the 4 seconds and can't figure out where it fits in.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 14, 2007 #2

    Hootenanny

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    You need to start by calculating the acceleration of the masses.
     
  4. Oct 14, 2007 #3
    I see 3 options

    1. Use suvat
    2. Final V - Initial V/Total time (I don't know any velocities)
    3. Drop AS Physics


    I'm lost
     
  5. Oct 14, 2007 #4

    Hootenanny

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    I see a fourth option;

    4. Newton's second law
     
  6. Oct 14, 2007 #5
    Ok, now I have the mass, but not the acceleration or the force. Is this where I use gravity?
     
  7. Oct 14, 2007 #6

    Hootenanny

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    Indeed, you need to examine the forces acting on each mass. I suggest you consider each mass as a seperate entity and draw a seperate FBD for each.
     
  8. Oct 14, 2007 #7
    Ok, so I started with the 4kg weight and got a 40N force as gravity, or am I doing something wrong here
     
  9. Oct 14, 2007 #8

    Hootenanny

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    Nope, that looks good to me; but is weight the only force acting here?
     
  10. Oct 14, 2007 #9
    Hmm, so I was thinking that the other force would be the 6kg weight pulling downwards at 60N, do I take the leftover 20N and put that into f=ma?
     
  11. Oct 14, 2007 #10

    Hootenanny

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    Whoa... slow down there. I said that we should consider each mass seperately, so whats the other force acting on the 40kg mass besides it's weight?
     
  12. Oct 14, 2007 #11
    Damn, I thought I was on to something there.

    I really have no idea about the other force now, it's probably something really basic that I passed over
     
  13. Oct 14, 2007 #12

    Hootenanny

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    Okay, how about the tension in the string?
     
  14. Oct 14, 2007 #13
    Ah, right. I can't remember covering tension in my Physics class yet, though I seem to make a connection with a formula including change in length and original length, I'm not sure though.
     
  15. Oct 14, 2007 #14

    Hootenanny

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    We're not really bothered about how the tension affects the string (we're assuming that the string inextensible), all we're really interested in is how it affects the mass. So, there are two forces acting on the 40kg mass, tension upwards and its weight downwards. This means we can write an expression using Newton's second law;

    [tex]F_{net}=ma[/tex]

    [tex]T - mg = ma[/tex]

    [tex]T - 40g = 40a[/tex]

    Do you follow?
     
  16. Oct 14, 2007 #15
    So the tension is going to reduce the resultant force that causes the acceleration. I still can't see how I get the f=ma to work here though, T-40g=40a still leaves me with tension and acceleration unsolved :S What did I miss
     
  17. Oct 14, 2007 #16

    Hootenanny

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    Well, whats say we have a look at the 60kg mass and do the same...?
     
  18. Oct 14, 2007 #17
    So we'd end up with T-60g=60a and T-40g=40a? I tried making a simultaneous equation here and ended up with T=240a and a=2 for the 40kg mass, is this good to go or did I miss the boat?

    Also, shouldn't it be 4kg and 6kg instead of 40 and 60 (unless there was a reason that i missed)
     
  19. Oct 14, 2007 #18
    Anybody still online who can help me out?
     
  20. Oct 14, 2007 #19

    Hootenanny

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    Careful here, you need to examine the direction of the forces. Is the tension really acting in the same direction that the 6kg mass is moving?
    Yes, sorry my bad, it should of course be 4kg and 6kg.
     
  21. Oct 15, 2007 #20
    So T+60=6a and T-40=4a, was I right about the simultaneous eq'n?
     
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