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19 October, 02:57

Find the network done by friction on a box that moves in a complete circle of radius 1.82 m on a uniform horizontal floor. The coefficient of kinetic friction between the floor and the box is 0.25, and the box weighs 65.0 N. A) - 0 B) - 370JC) - 190 J D-1800 J

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Answers (2)
  1. 19 October, 03:34
    0
    C) W = - 190 J

    Explanation:

    Notation

    Wf = work done by the friction force (unknown)

    Ff = force of the friction

    d = distance travelled by the box = (2 pi 1.82 m) = 11.435 m
  2. 19 October, 05:40
    0
    the network done by friction on a box that moves in a complete circle is 185.7 joules

    Explanation:

    Step one

    Given

    Radius of circle = 1.82m

    Circumference of the circle = 2*pi*r

    =2*3.142*1.82=11.43

    Hence distance = 11.43m

    Coefficient of friction u=0.25

    Weight of box = 65N

    We know that work = force*distance

    But the limiting force = u*weight

    Hence the net work done by friction

    Wd=0.25*65*11.43

    Wd=185.7 joules
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