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I don't get this?

  1. Jan 9, 2005 #1
    I need help to figure this problem out. I don't kno what formula to use to solve this!

    A shopper in a supermarket pushes a loaded 32kg cart with a horizontal force of 12N. How far will the cart move in 3.5s if the shopper places an 85N child in the cart before pushing.

    I only need the formula so plz help!
     
  2. jcsd
  3. Jan 9, 2005 #2
    when they use newtons in terms of weight divide N = mg so 85N/9.8 gives the mass, with that mass add it to the mass of the cart. divide the 12N force by the total mass to find the acceleration. with the acceleration input into formula (Vi)t + .5at^2 = d Vi = zero because the cart is not moving initially therefore velocity = zero. so .5at is left so .5 multiplied by the acceleration you found multiplied by time squared, 3.5s^2
     
    Last edited: Jan 9, 2005
  4. Jan 9, 2005 #3

    dextercioby

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    U need to compute the total mass of the cart.Newton's law (theII-nd) says that
    [tex] F=ma [/tex]
    Once u have found the acceleration,u can use this formula
    [tex] s=\frac{1}{2}at^{2} [/tex]

    Daniel.
     
  5. Jan 9, 2005 #4

    dextercioby

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    Wrong.The formula includes a product between the acceleration and the time SQUARED.

    Daniel.
     
  6. Jan 9, 2005 #5
    Yes I corrected it
     
  7. Jan 9, 2005 #6
    i still don't get it since the answer is supposed to be 2.3M and i get 20m
     
  8. Jan 9, 2005 #7
    I get 40.7 for total mass 12/40.7(.5)(3.5^2)=1.8M, i get it closer but its still wrong
     
  9. Jan 9, 2005 #8

    dextercioby

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    Both answers look wrong to me...What did u get for the mass??What about acceleration??

    Daniel.
     
  10. Jan 9, 2005 #9

    dextercioby

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    I basically did the same calculations.Why one earth would it yield different results??Is there something we missed...?? :confused:

    Daniel.
     
  11. Jan 9, 2005 #10
    for mass i got 40.67kg for mass and 3.39 for acceleration
     
  12. Jan 9, 2005 #11
    How did u get 3.39? If mass is 40.67 shouldnt acceleration be 12/40.67?
     
  13. Jan 9, 2005 #12
    how did u guyz get 1.8m i get 20m when i do those calculations
     
  14. Jan 9, 2005 #13

    dextercioby

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    The mass is good,the acceleration is wrong.Wheck your numbers once again.

    Daniel.
     
  15. Jan 9, 2005 #14
    LoL, is that cart driver throwing the child in the cart, thus creating intitial velocity?

    Is there anything else to the problem or is this it?

    And are you sure the answer is suppsoed to be 2.3M?
     
    Last edited: Jan 9, 2005
  16. Jan 9, 2005 #15
    ok thats it nvm thanx i kinda also need help on a kinetic friction problem. i am not good with formulas so i don't kno how to solve for this either.

    A 4kg block is pushed alomng the ceiling with a constant apllied force of 85N that acts at an angle of 55 degrees with the horizontal. The block accelerates to the right at 6m/s^2. detremine the coefficient of kinetic friction between the block and the ceiling.

    the answer is supposed to be 0.816 plz help and thanx again for the last one
     
  17. Jan 10, 2005 #16

    dextercioby

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    Do you know the expression for the kinetic friction force??

    Daniel.
     
  18. Jan 10, 2005 #17
    umm i only kno that stuff in the question so i have no clue how to solve
     
  19. Jan 10, 2005 #18

    dextercioby

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    Then how would you go about solving it...??I will not solve it for you.It's not the plicy here.We only give indications.
    HINT:
    [tex] |\vec{F}_{fr.Kin.}|=\mu_{Kin} |\vec{N}| [/tex]

    Can u take it from here??
    Do u know how to apply Newton's laws??

    Daniel.
     
  20. Jan 10, 2005 #19
    Well you have to find the vertical component of the force using trig, sine will work. That will be your "normal" force. so Fn(CoeFriction)=Ff Ff is the force that friction is exerting in the opposite direction of movment. Take it from there
     
  21. Jan 10, 2005 #20
    i don't want u guyz to solve it for me of course but i don't understand what either of u are trying to tell me
     
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