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mario62 [17]
3 years ago
13

An 85 kg clock initially at rest on a horizontal floor requires a 680 N horizontal force to set it in motion. After the clock is

in motion, a horizontal force of 540 N keeps it moving with a constant velocity. Find μs and μk between the clock and the floor.
Physics
1 answer:
Aleks04 [339]3 years ago
8 0

A force of 680 N is required to get the clock moving, so the maximum static friction is also <em>f</em> ˢ = 680 N. The clock is at rest, so the net vertical force acting on it is 0, and by Newton's second law,

<em>n</em> - <em>mg</em> = 0

where

<em>n</em> = magnitude of the normal force

<em>m</em> = 85 kg = mass of the clock

<em>g</em> = 9.8 m/s² = magnitude of the acceleration due to gravity

So we have

<em>n</em> = <em>mg</em> = (85 kg) (9.8 m/s²) = 833 N

which means the static friction <em>f</em> ˢ is such that

<em>f</em> ˢ = <em>µ </em>ˢ <em>n</em>

Solving for the coefficient of static friction gives

<em>µ</em> ˢ = (680 N) / (833 N) ≈ 0.82

After it starts moving, a force of 540 N is required to keep the clock going at a constant speed, so the kinetic friction is also <em>f</em> ᵏ = 540 N. Then

<em>f</em> ᵏ = <em>µ</em> ᵏ <em>n</em>

and solving for the coefficient of kinetic friction yields

<em>µ</em> ᵏ = (540 N) / (833 N) ≈ 0.65

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