Question: Some sliding rocks approach the base of a hill with a speed of . The hill rises at 36° above the horizontal and has coefficients of kinetic friction and static friction of 0.45 and 0.65, respectively, with these rocks. (a) Find the acceleration of the rocks as they slide up the hill. (b) Once a rock reaches its highest point, will it stay there or slide down the hill? If it stays, show why. If it slides, find its acceleration on the way down.

• The speed of the rock is, $v=12\text{m/s}$.
• The angle of the hill is, $\theta =36°$.
• The coefficient of kinetic friction is, ${\mu }_k=0.45$.
• The coefficient of static friction is, ${\mu }_s=0.65$.

The coefficient of friction is the amount of friction that occurs when one surface slides over another surface. The less value of coefficient of friction indicates that less force is required for sliding a surface over another surface.

From the free body diagram as shown above the expression for force at y-axis can be expressed as,

$$\begin{gathered} \sum F_y=0 \\ {R}_N-mg\cos \theta =0 \\ {R}_N=mg\cos \theta \end{gathered}$$

Here$R_N$ is the normal reaction force, m is the mass of the rock, g is the acceleration due to gravity.

The expression for the force at x-axis can be expressed as,

$$\begin{gathered} \sum F_x=ma \\ -mg\sin \theta -f_k=ma \\ -mg\sin \theta -{\mu }_k{R}_N=ma \end{gathered}$$

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

Substitute $mg\cos \theta$for $R_N$in the above equation.

$$\begin{gathered} -mg\sin \theta -{\mu }_k(mg\cos \theta )=ma \\ a=-g(\sin \theta +{\mu }_k\cos \theta ) \end{gathered}$$

Substitute $9.8m/s^2$for g, $36°$for $\theta$, and 0.45 for ${\mu }_k$in the above equation.

$$\begin{gathered} a=-(9.8{\text{m/s}}^{\text{2}})(\sin 36°+0.45\cos 36°) \\ a=-9.32{\text{m/s}}^{\text{2}} \end{gathered}$$

Hence, the required acceleration is, $-9.32m/s^2$.

From the free body diagram the force on y-axis can be expressed as,

$$\begin{gathered} \sum F_y=0 \\ {R}_N-mg\cos \theta =0 \\ {R}_N=mg\cos \theta \end{gathered}$$

From the free body diagram the force on x-axis can be expressed as,

$$\begin{gathered} \sum F_x=-ma \\ -mg\sin \theta +f_k=ma \\ mg\sin \theta -{\mu }_k{R}_N=ma \end{gathered}$$

Substitute $mg\cos \theta$for $R_N$in the above equation.

$$\begin{gathered} mg\sin \theta -{\mu }_k(mg\cos \theta )=ma \\ a=g(\sin \theta -{\mu }_k\cos \theta ) \end{gathered}$$

Substitute $9.8m/s^2$for g,$36°$ for $\theta$, and$0.45$ for${\mu }_k$ in the above equation.

$$\begin{gathered} a=(9.8{\text{m/s}}^{\text{2}})(\sin 36°-0.45\cos 36°) \\ a=2.19{\text{m/s}}^{\text{2}} \end{gathered}$$

Hence, the required acceleration is, $2.19m/s^2$.