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