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    can someone give me a hand with this question i am completely lost!

    A box of mass 5kg is being pulled along a rough horizontal ground by a rope inclined at 20 degrees to the horizontal. When the tension in the rope is 10 N the box is moving at a constant speed.

    A) calculate the coefficient of friction between the box and the ground.

    b) The tension in the rope is now 15 N
    calculate the acceleration of the box
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    10N = limiting friction and weight acting down the slope
    10 = F + 5gsin20
    F = 10 - 5gsin20
    normal reaction = 5gcos20
    (mu) = (10 - 5gsin20)/5gcos20

    F = ma
    15 - (10 - 5gsin20) - 5gsin20 = 5a
    15-10 = 5a
    a = 1 ms-2
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    (Original post by Jutx)
    can someone give me a hand with this question i am completely lost!

    A box of mass 5kg is being pulled along a rough horizontal ground by a rope inclined at 20 degrees to the horizontal. When the tension in the rope is 10 N the box is moving at a constant speed.

    A) calculate the coefficient of friction between the box and the ground.

    b) The tension in the rope is now 15 N
    calculate the acceleration of the box
    Resolve horizontally:
    10cos20-UR=0
    10cos20-5gU=
    5gU=10cos20
    U=10cos20/5g
    =10cos20/49
    =0.192

    b) 15cos20- (0.192x49)=5a
    4.687=5a
    a=0.94ms-2
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    cheers. I am not the best person at mechanics lol
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    (Original post by Undry1)
    10N = limiting friction and weight acting down the slope
    10 = F + 5gsin20
    F = 10 - 5gsin20
    normal reaction = 5gcos20
    (mu) = (10 - 5gsin20)/5gcos20

    F = ma
    15 - (10 - 5gsin20) - 5gsin20 = 5a
    15-10 = 5a
    a = 1 ms-2
    Horziontal rough plane
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    (Original post by Malik)
    Horziontal rough plane
    oops.
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    (Original post by Jutx)
    can someone give me a hand with this question i am completely lost!

    A box of mass 5kg is being pulled along a rough horizontal ground by a rope inclined at 20 degrees to the horizontal. When the tension in the rope is 10 N the box is moving at a constant speed.

    A) calculate the coefficient of friction between the box and the ground.

    b) The tension in the rope is now 15 N
    calculate the acceleration of the box
    R = upwards force = 10sin20 + s

    vertically: R = 10sin20 + s = 5g
    horizontally: F = µR = 10cos20

    µ = 10cos20/5g = 0.19...

    b) F = ma
    a = F/m = (15cos20 - µ5g)/5 = 0.93ms^-2 (which is cos20, don't you know)
 
 
 
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