91Ó°ÊÓ

The given data can be listed below as,

• The mass of box of textbook is 25.0kg.
• The coefficient of kinetic friction is ${\mathrm{μ}}_k=0.25$.
• The coefficient of static friction is ${\mathrm{μ}}_s=0.35$.
• Distance of box when slide along the ramp is s = 5.0m.

The kinetic friction is the force of friction that occurs when the object is accelerating and the static friction is the force of friction that occurs when the object is at rest.

In the free body diagram given above, the all forces act on the box can be expressed as,

$mg+F_{fr}+N=0$

Here, m is the mass of the box, g is the acceleration due to gravity and $F_{fr}$is the frictional force and N is the normal force.

The projected forces of x and y axis are expressed as,

$$\begin{gathered} mg\sin \mathrm{Î\pm }-F_{fr}=0 \\ N-mg\cos \mathrm{Î\pm }=0 \\ N=mg\cos \mathrm{Î\pm } \end{gathered}$$

From the above free body diagram the frictional force act on the box is expressed as,

$\begin{array}{rcl}{F}_{fr}& =& {\mathrm{μ}}_{s}N\\ & =& {\mathrm{μ}}_{s}mg\cos \mathrm{Î\pm }\end{array}$

Here,${\mathrm{μ}}_s$ is the coefficient of static friction.

Substitute the value of $F_{fr}$and 0.35 for ${\mathrm{μ}}_s$in the above equation.

$$\begin{gathered} mg\sin \mathrm{Î\pm }-{\mathrm{μ}}_{s}mg\cos \mathrm{Î\pm } \\ \sin \mathrm{Î\pm }={\mathrm{μ}}_s\cos \mathrm{Î\pm } \\ \tan \mathrm{Î\pm }=0.35 \\ \mathrm{Î\pm }={\text{tan}}^{-1}(0.35) \end{gathered}$$

$\mathrm{Î\pm }=19.29°$

Hence, the required angle is $19.29°$.

The projection of the forces on x axis can be expressed as,

$$\begin{gathered} mg\sin \mathrm{Î\pm }-{\mathrm{μ}}_{k}mg\cos \mathrm{Î\pm }=ma \\ mg\left(\sin \mathrm{Î\pm }-{\mathrm{μ}}_k\cos \mathrm{Î\pm }\right)=ma \\ a=g\left(\sin \mathrm{Î\pm }-{\mathrm{μ}}_k\cos \mathrm{Î\pm }\right) \end{gathered}$$

Here, a is the acceleration, ${\mathrm{μ}}_k$is the coefficient of kinetic friction.

Substitute $9.8m/s^2$for g, $19.29°$for $\mathrm{Î\pm }$and 0.25 for ${\mathrm{μ}}_k$in the above equation.

localid="1668414612714" $\begin{array}{rcl}a& =& \left(9.8m/s^2\right)(\sin 19.29°-0.25×\cos 19.29°)\\ & =& 0.931m/s^2\end{array}$

Hence, the required acceleration is $0.931m/s^2$.

The kinematic equation of the velocity is expressed as,

$$\begin{gathered} v^2=v_0^2+2as \\ v=\sqrt{v_0^2+2as} \end{gathered}$$

Here, v is the velocity of box after slide, $v_0$is the initial velocity and a is the acceleration and s is the sliding distance.

Substitute 0m/s for $v_0$, $0.931m/s^2$for a, and 5.0m for s in the above equation.

$\begin{array}{rcl}v& =& \sqrt{0+2(0.931m/s^2)(5.0m)}\\ & =& 3.05m/s\end{array}$

Hence, the required velocity is 3.05m/s.