$\begin{array}{1 1}0.48\\0.36\\0.24 \\0.62\end{array} $

Step 1:

Draw the free body diagrams :

The box is being moved up :

The box is being moved down :

Step 2 :

The force $F_G$ decomposes into two components in the direction of these axes :

$F_{Gx} =F_G \sin \alpha$

$F_{Gy} =F_G \sin \alpha$

The box is moved up :

$x: -F_G \sin \alpha +F_1- F_t=0$ ----(1)

$y : N- F_G \cos \alpha =0$-----(2)

The box is moved down :

$x: -F_G \sin \alpha +F_2- F_t=0$ ----(3)

$y : N - F_G \cos \alpha =0$-----(2)

Step 3:

The friction force that affects the box is directly proportional to the normal force by which the box affects the inclined plane .

This force is , according to Newton's third law, of the same magnitude as the force by which the inclined plane affects the box .

The friction force$F_t$ can therefore be expressed as follows :

$F_t= \mu _k N$ ------(4)

We express the force $F_t$ can therefore be expressed as follows :

We express the force N from equation (2) :

$N= F_G \cos \alpha$

Equation for the friction force :

$F_t = \mu _k F_G \cos \alpha$ -------(5)

Step 4 :

Determining the coefficient of kinetic friction $\mu_K$

We substitute into motion equation (1) and (3) the equation for the friction force (5).

We express the force $F_1 $ as $6 F_2$

$-F_G \sin \alpha +6 F_2- F_G \mu _K \cos \alpha =0$ -------(6)

$F_G \sin \alpha + F_2 -F_G \mu _K \cos \alpha =0$ -------(7)

Equation (7) is multiplied by 6 and subtracted from equation (6):

$7 F_G \sin \alpha - 5 F_G \mu _k \cos \alpha =0$

$7F_G \sin \alpha \alpha = 5 F_G \mu _K \cos \alpha /: (5 \;F_G \cos \alpha)$

$\mu_K =\large \frac{7}{5} $$ \tan \alpha$---(8)

For the given numerical value we get from (8):

$\mu_k =\large\frac{7}{5} $$ \tan 15^{\circ}$

$\qquad= \large\frac{7}{5}$$. 0.258$

$\qquad= 0.36 $

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