Question

A 25.0-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?

Answer 1

Answer:

(A) Frictional force will be 103.7276 N

(B) Coefficient of friction $\mu _k=0.4751$

Explanation:

We have given

mass of the box m = 25 kg

Angle $\Theta =27^{\circ}[/text]Acceleration [tex]a=0.3m/sec^2$

Apply Newton's second law of motion ,

Vertically :

            $F_N$= mg cosθ

                = $25\times 9.8\times COS27^{\circ}=218.296N$

Horizontally :

              mg sinθ - F = ma

A)

frictional force F= mg sinθ - ma

                        = $25\times 9.8sin27^{\circ}-25\times 0.3=103.7276N$

B)

Coefficient of friction  $\mu _k=\frac{F}{F_N}=\frac{103.7276}{218.296}=0.4751$