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The 1kg box shown in the figure has a force of f=21 N applied at an...

The 1kg box shown in the figure has a force of f=21 N applied at an angle of theta = 60 degrees this box travels on a distance of d=20 M. The coefficient of kinetic friction between the box and the surface is 0.2 to the nearest joule, what is the magnitude of the work done by friction on this box?
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Answer #1

When the block on an inclined plane than the normal reaction is

N = mgcos60

= 1*9.81*cos60

= 4.905 N

Calculate the magnitude of the friction force between the box and the surface .

Fr = µ*N

    = 0.2* 4.905

   = 0.981 N

Calculate the work done by the friction.

W = Fr*d

    = 0.981 *20

   = 19.62 J

Thus, the work done by the friction is 19.62 J .

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