When an object is placed on an incline which is not frictionless, there are two forces acting on it. One is the force due to the gravitational pull of the Earth, in the downward direction and the other is an opposing force due to friction.

In the problem, the object does not have any acceleration. This means that the two forces acting on it are equal.

The gravitational force downwards can be expressed as a sum of two vectors, one normal to the inclined plane and the other parallel to the plane.

The magnitude of the component normal to the plane is m*g*cos 47.9, where m is the mass and g is the acceleration due to gravity.

The magnitude of the component parallel to the plane and acting in a downwards direction is m*g*sin 47.9.

The force of kinetic friction opposing any acceleration due to the the force of gravitation is N*Kf, where N is the normal force and Kf is the coefficient of kinetic friction.

N*Kf = m*g*sin 47.9

=> m*g*cos 47.9*Kf = m*g*sin 47.9

=> Kf = sin 47.9 / cos 47.9

=> Kf = tan 47.9

=> Kf = 1.106

**The required coefficient of kinetic friction is 1.106**