Verified answer

At 34.4° the block starts to slide down the incline

At 34.4° the component of weight down slope = mg sin 34.4°

and the component of weight perpendicular to the slope = mg cos 34.4°

Now at the point that the block just starts to move the static friction is greatest and F(friction) = µ(static) N = µ(static) . mg cos 34.4°

This greatest static friction just equals the component of weight down the slope

So µ(static) . mg cos 34.4° = mg sin 34.4°

µ(static) = sin 34.4° / cos 34.4° = tan 34.4° [the mg's cancel]

µ(static) = 0.6847 [To 4 dec pl]

So the coefficient of static friction is 0.6847

Now when the block moves at constant speed down the slope for some angle you can calculate the coefficient of kinetic friction as tan (that angle)

But we're not told that the speed it has is constant

The block travels 3.30m in 2.20 s so it's velocity down the slope = 3.30/2.20 = 1.5m/s

So the acceleration of the block down the slope = (1.5 - 0) / 2.20 = 0.68 m/s² [Because it starts at rest]

The resultant force down the slope = ma = 0.68m

So 0.68 m = mg sin 34.4° - µ(kinetic) . mg cos 34.4°

0.68 = 9.8 sin 34.4° - µ(kinetic) . 9.8cos 34.4°

µ(kinetic) . 9.8 cos 34.4° = 9.8 sin 34.4° - 0.68

µ(kinetic) = (9.8 sin 34.4° - 0.68) / 9.8 cos 34.4°

µ(kinetic) = 0.6006 [To 4 dec pl]

So the coefficient of kinetic friction is 0.6006

[µ(kinetic) is always less than µ(static)]