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Find the net work done by friction on a box that moves in a complete circle of radius 0.40m on a uniform horizontal floor. The coefficient of kinetic friction between the floor and the box is 0.13, and the box weights 35.0N

Respuesta :

Answer:

42/ 5 pi J

Explanation:

The word done by friction is given by

[tex]F=\mu Nx[/tex]

where μ = coefficent of frction, N = weight of the box, and x = distance traveled.

Now in our case

μ = 0.13

N = 35.0 N

x = 0.40 * 2 pi (since it is the circumeference of the circle

Therefore,

[tex]F=(0.13)(35.0)(0.40\cdot2\pi)[/tex]

which simplifies to give

[tex]F=\frac{42}{5}\pi[/tex]

which is our answer!

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