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vaieri [72.5K]
3 years ago
7

A 25 kg box is being pulled at a constant velocity with a tension force of 65 N. what is the coefficient of friction between the

box and the ground?
Physics
1 answer:
DiKsa [7]3 years ago
7 0

The box is in equilibrium, so Newton's second law says

<em>n</em> + (-<em>w</em>) = 0

65 N + (-<em>f</em> ) = 0

where <em>n</em> denotes the magnitude of the normal force, <em>w</em> denotes the weight of the box, and <em>f</em> denotes the magnitude of the friction force.

The box has a weight of

<em>w</em> = (25 kg) (9.80 m/s²) = 245 N

so <em>n</em> = 245 N, too.

The friction force has magnitude

<em>f</em> = 65 N

and is proportional to the normal force by a factor of <em>µ</em>, the coefficient of kinetic friction. So we have

65 N = <em>µ</em> (245 N)   →  <em>µ</em> ≈ 0.26

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F=27.39N

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