A force of 680 N is required to get the clock moving, so the maximum static friction is also f ˢ = 680 N. The clock is at rest, so the net vertical force acting on it is 0, and by Newton's second law,
n - mg = 0
where
n = magnitude of the normal force
m = 85 kg = mass of the clock
g = 9.8 m/s² = magnitude of the acceleration due to gravity
So we have
n = mg = (85 kg) (9.8 m/s²) = 833 N
which means the static friction f ˢ is such that
f ˢ = µ ˢ n
Solving for the coefficient of static friction gives
µ 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 f ᵏ = 540 N. Then
f ᵏ = µ ᵏ n
and solving for the coefficient of kinetic friction yields
µ ᵏ = (540 N) / (833 N) ≈ 0.65