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A 6.00 kg crate is pushed by a 300.0 N force at and angle of 30 degrees with the horizontal.
a.What coefficient of friction value is experienced by the crate if it moves at constant velocity?
b. What coefficient of friction value is present if the crate accelerates at 1.50m/s2

The answer to (a) depends upon whether the 30 degree angle of the pushing force is upwards or downwards, with respect to the floor. I will assume downwards.

The weight of the crate is M g = 6.0 x 9.8 = 58.8 N

The total force component downwards on the floor is then 58.8 + 300 sin 30 = 208.8 N

If there is no acceleration, the net horizontal force is zero

300 cos 30 - 208.8 * f = 0
Solve for the coefficient of friction, f

For (b), get the net force from
Fnet = M a = 450 N
Fnet = 300 cos 30 - 208.8 f = 450
Solve for the (new) value of f.

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