Question

Some sliding rocks approach the base of a hill with a speed of 12 m/s. 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.

Short answer

The answer is not given in the drive.

Step-by-step solution

Identification of given data

• The speed of the rock is, v = 12 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, ${\mu }_k=0.65$.

Significance of coefficient of friction

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.

Determination of acceleration as rock slid up the hill.

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 m g cos$\theta$ for$R_N$ in the above equation.

role="math" localid="1667635448075" $$\begin{gathered} -mg\sin \theta -{\mu }_k(mg\cos \theta )=ma \\ a=-g\left(\sin \theta -{\mu }_k\cos \theta \right) \end{gathered}$$

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

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

Hence, the required acceleration is,$-9.32\mathrm{m}/{\mathrm{s}}^2$ .

Determination of acceleration on the way down.

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 m g cos$\theta$for$R_N$in the above equation.

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

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

localid="1667642664203" $$\begin{gathered} \mathrm{a}=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\sin 36°-0.45\cos 36°\right) \\ \mathrm{a}=2.19\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the required acceleration is,$2.19\mathrm{m}/{\mathrm{s}}^2$.