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Thread: physicis problem #2

  1. #1
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    physicis problem #2

    1) A box of books weighing 200 N is shoved across the floor of an apartment by a force of 300 N exerted horixontally.
    if the cofficient of kinetic friction between the floor anf box is 0.52, how long does it take the box th move 3.0m starting fron rest

    I want the concepts to solve this queation
    Last edited by mr fantastic; Nov 5th 2010 at 01:04 PM. Reason: Title.
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  2. #2
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    Quote Originally Posted by r-soy View Post
    1) A box of books weighing 200 N is shoved across the floor of an apartment by a force of 300 N exerted horixontally.
    if the cofficient of kinetic friction between the floor anf box is 0.52, how long does it take the box th move 3.0m starting fron rest

    I want the concepts to solve this queation
    Let F = applied force of 300 N

    f = force of kinetic friction

     N = Normal force acting on the box

    W = weight of the box (200 N)

    (note that N = W in this case)

     \mu = coefficient of friction (0.52)

     m = mass of the box (not the weight)

     a = acceleration of the box


    F_{net} = F-f

    ma = F - \mu \cdot N

    ma = F - \mu \cdot W

    \displaystyle a = \frac{F - \mu \cdot W}{m}

    once you've calculated the acceleration ...

    \Delta x = \frac{1}{2} at^2 ... solve for t
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  3. #3
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    a = 300-(0.52)(200)/400
    = 299

    a=1/2(299)t^2
    t = 0.04

    Is ok ?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    No. Could you show us how you got the mass of the box to be 400 kg?

    Later, you should get the time as 2.5 s.


    And your calculation for:
    a=1/2(299)t^2
    t = 0.04

    Is not good either.

    \Delta x = \dfrac12 at^2

    30 = \dfrac12 (299)t^2

    t^2 = \dfrac{30 \times 2}{299} = 0.2

    t = \sqrt{0.2}= 0.45 s
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  5. #5
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    Is the answer now
    0.45 ??
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  6. #6
    MHF Contributor Unknown008's Avatar
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    No, the 0.45 s was the reply concerning your wrong calculation so that you know where you did a wrong calculation so that next time, you know what to do using the correct values.

    I told you to show us how you got 400 kg for the mass.
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  7. #7
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    How i can find the mass ??

    i find mass 2 X 200 = 400
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  8. #8
    MHF Contributor Unknown008's Avatar
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    And can I know why you did that?

    Let me remind you that:

    W = mg

    W is the weight, m is the mass and g the acceleration due to gravity.
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  9. #9
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    a = 300-(0.52)(200)/20
    =9.8

    t^2 = 30 X 2 /9.8 = 6.12
    t = 2.4

    now is ok ??
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  10. #10
    MHF Contributor Unknown008's Avatar
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    Right, but you usually have to round it off well. All the numbers are 2.4743583 which gives 2.47 or 2.5 seconds.

    I thought you would have taken g as 9.8 m/s^2... but nevermind. If you are told to use g = 10m/s^2, you use it as such.
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