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Law of motion problem

  1. Jul 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Q69
    20150722_173223.jpg

    2. Relevant equations
    F=ma
    v = u + at
    v^2 = u^2 + 2as
    s = ut + (1/2)at^2

    3. The attempt at a solution
    Find the deceleration by friction:
    F = ma
    0.1 = (0.2)a
    a = 0.5 ms^-2

    and the acceleration of the mug would be 3 - 0.5 = 2.5 ms^-2

    Find the time required for the tablecloth to travel 30 cm
    u = 0, , s = 0.3, a = 3, t = ?
    s = ut + (1/2)at^2
    0.3 = (1/2)(3)(t^2)
    t = 0.447 s

    Find the displacement of mug within 0.447s
    t = 0.447, s = ?, u = 0, a = 2.5
    s = ut + (1/2)at^2
    s = (1/2)(2.5)(0.2)
    s= 0.25 m , but the answer is 0.06 m

    Thanks!
     
  2. jcsd
  3. Jul 22, 2015 #2

    jbriggs444

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    In which direction is the mug accelerating? In which direction does friction between mug and tablecloth act? And what, exactly, do you mean by "a" here?
     
  4. Jul 22, 2015 #3
    Is the acceleration of block 2.5m/s-2 in ground frame ?

    Hope this helps .
     
  5. Jul 22, 2015 #4

    jbriggs444

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    If you draw a free body diagram for the block, what forces act on it?
     
  6. Jul 22, 2015 #5
    2.5m/s-2 is acceleration of block relative to cloth .
     
  7. Jul 22, 2015 #6

    jbriggs444

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    Good. And its acceleration relative to the ground is?
     
  8. Jul 22, 2015 #7
    0.5m/s-2 - Thanks jbriggs444 .
     
    Last edited: Jul 22, 2015
  9. Jul 23, 2015 #8
    why is the acceleration of the mug relative to the ground not (2.5 + 3) because the acceleration of the tablecloth is 3 and the mug is sitting on it and accelerate with 2.5 ms^2, so the acceleration of the mug relative to the ground should be 5.5 ms^2

    but even if I use a = 0.5 to calculate, I still cannot get the answer. Why's that?
     
    Last edited: Jul 23, 2015
  10. Jul 23, 2015 #9
    Okay - you seem to be confused .

    Try answering these questions :

    then

    and then

     
  11. Jul 23, 2015 #10
    Use a = 0.5m/s2 only in your last step - you will get your answer .

     
  12. Jul 23, 2015 #11
    Thanks, I have got the answer!

    But if I want to make the calculation process easier (without the free body diagram), can I just think it in this way?
    Since the friction between the mug and the tablecloth is 0.1N, meaning that the force transferred from the tablecloth to the mug for the mug to move can only be 0.1N, so I should

    F = ma
    0.1 = 0.2a
    a = 0.5 ms^-2
     
  13. Jul 23, 2015 #12
    Force is not transferred , it is exerted . And no , you don't need the FBD for this question - it was only for you to visualize a bit better .

    The rest is fine .
     
    Last edited: Jul 23, 2015
  14. Jul 29, 2015 #13
    Your working is wrong.
    The correct solution is:
    mass of mug = 200g = 0.2kg
    mug located = 30cm = 0.3m (from the edge of the table)
    Acceleration of cloth = 3m/s^2
    Using s = ut + 1/2at^2
    Finding time:
    0.3 = 0+1/2*3*t^2
    t = 0.2 sec

    Finding the how far mug can move.
    s = ut + 1/2at^2
    s = 0 + 1/2x3x(0.2)^2
    s = 0.06m

    The answer is 0.06m.
     
  15. Jul 29, 2015 #14
    You've made a mistake here - t2 = 0.2 sec .

    Also , try to not solve the entire question - You can read about this in the guidelines .
     
  16. Jul 29, 2015 #15

    haruspex

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    That's completely wrong.
    First, the given distance is not to the edge of the table, it's to the edge of cloth.
    Second, your calculation of the time would find how long it takes for the cloth to move .3m. This does not tell you where the mug is in relation to the cloth at that time.
    Third, you then used the acceleration of the cloth as though it were the acceleration of the mug. The only reason you did not end up with an answer of 0.3m is your fourth error, confusing t with t2, as Qwertywerty noted.

    Fwiw, the quickest way is to note that the acceleration of the mug is 1/6 that of the cloth. Since t2 is the same for both, it travels 1/6 of the distance. The difference between the two distances is 0.3m.
     
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