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Newton's Second law

  1. Oct 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Influenced by certain force, cart moved from place and driven 0.5 meters distance. When 200grams weight was placed on cart it moved 0.3 meters during the same period of time. Calculate the mass of the cart.
     
  2. jcsd
  3. Oct 10, 2015 #2
    So, how did you tried?
     
  4. Oct 10, 2015 #3
    Tried to get acceleration by converting grams to kilos, then dividing 0,3 meters with 0,2 kilos. Got 1,5, but I doubt its right. If 1,5 would be correct, I would just use ΣF⃗=ma⃗
     
  5. Oct 10, 2015 #4
    Then, why do you need the acceleration? I know you want to use the Newton's second law, but there are many missing parameters to use this.
     
  6. Oct 10, 2015 #5
    I want acceleration to get force and with Force + Acceleration I would get mass. So what I should use instead? Pretty lost,
     
  7. Oct 10, 2015 #6
    In this statement, I can get the velocity of cart and cart + 200grams weight, roughly. But I don't know how to get the acceleration.
     
  8. Oct 10, 2015 #7
    How about a1=V1-V01/t a2=V2-V02/t that leads to (V1-V01)*t/t*(V2-V02), we cancel out t and we get (V1-V01)/(V2-V02), now I need to know how you get velocity of cart and cart + 200grams.
     
  9. Oct 10, 2015 #8

    ehild

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    You know the displacements. Assuming uniform acceleration, how is the displacement related to time and the acceleration?
     
  10. Oct 10, 2015 #9
    Displacement moves same distance each time interval
     
  11. Oct 10, 2015 #10
    What you means is the constant velocity. Is this problem settle down with constant acceleration or constant velocity?
     
  12. Oct 10, 2015 #11
    You mentioned you are abble to get velocity, if you would tell me how, I (maybe) would be abble to get acceleration from there and use it to get mass.
     
  13. Oct 10, 2015 #12
    I supposed the cart moves without acceleration, that is constant velocity. Since we know the displacement within unknown time interval ##T##, I can only tell you that the velocity is in a form ## 0.5 m / T ##.
     
  14. Oct 10, 2015 #13

    ehild

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    The problem statement is:
    If a certain force takes place, the velocity is not constant.
     
  15. Oct 10, 2015 #14
    I thought the certain force only acted at the initial and the cart moves freely. But, it seems you are right. My poor English sometimes confuse me.
     
  16. Oct 10, 2015 #15
    Ok, then now I am clear.

    The cart with mass ## M ## and cart + weight with mass ## M + m## moved by a fixed certain force ## F## and the displacements are ## d_1 ## and ## d_2 ##, respectively.
     
  17. Oct 10, 2015 #16

    ehild

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    That is true now.
    But this is Igniuxx6's thread. Do not confuse him, and do not try to solve his problem. Both are against the rules of these Forums.
     
  18. Oct 10, 2015 #17
    So
    m2=0,2kg
    m=m1+m2
    F=d1+d2/m2=0,5+0,3/0,2=4N
    a=F/m2=20
    m1=F/a=4/20=0,2kg
    m=0,2+0,2=0,4kg


     
  19. Oct 10, 2015 #18
    In constant acceleration, the displacement is well known and because we know the displacement, you can get the acceleration.

    This part is really strange. Why don't you check the relevant equations?
     
  20. Oct 10, 2015 #19
    So its like a=9,8?
    So it would be :
    F=m*a=0,2*9,8=1,96N
    m1=F/a=1,96/9,8=0,2kg
    m=0,2+0,2=0,4kg?
     
  21. Oct 10, 2015 #20
    For example, the displacement with constant velocity is given by ## d = v t. ## What about the displacement with constant acceleration?
     
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