A25.0-kg box is released on a 27° incline and accelerates down the incline at 0.30 m/s2. find the friction force impeding its motion. what is the coefficient of kinetic friction?

(A) Frictional force will be 103.7276 N

(B) Coefficient of friction

Explanation:

We have given

mass of the box m = 25 kg

Angle

Apply Newton's second law of motion ,

Vertically :

            = mg cosθ

                =

Horizontally :

              mg sinθ - F = ma

A)

frictional force F= mg sinθ - ma

                        =

B)

Coefficient of friction  

The frictional force impeding the motion of the box is 103.73 N

The coefficient of friction is 0.48

The given parameters;

The vertical component of the force on the box is calculated by applying Newton's second law of motion;

The frictional force is determined by resolving the horizontal component of the force on the box;

The coefficient of friction is calculate as;

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