The answer I'm going with is false
Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :

Horizontal component of force(Normal reaction) :

Since, box is on the verge of slipping :

Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
Answer and Explanation:
distance will be 2×3.14 (pie)×r
displacement will be 2r (diameter)
the motion is uniform circular motion as the object is moving in a circular path with uniform motion