91影视

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

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,

$$\begin{gathered} 鈭�F_y=0 \\ {R}_N-mg\cos 胃=0 \\ {R}_N=mg\cos 胃 \end{gathered}$$

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,

$$\begin{gathered} 鈭�F_x=ma \\ -mg\sin 胃-f_k=ma \\ -mg\sin 胃-{渭}_k{R}_N=ma \end{gathered}$$

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

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

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

Substitute $9.8m/s^2$for g, $36掳$for $胃$, and 0.45 for ${渭}_k$in the above equation.

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

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

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

$$\begin{gathered} 鈭�F_y=0 \\ {R}_N-mg\cos 胃=0 \\ {R}_N=mg\cos 胃 \end{gathered}$$

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

$$\begin{gathered} 鈭�F_x=-ma \\ -mg\sin 胃+f_k=ma \\ mg\sin 胃-{渭}_k{R}_N=ma \end{gathered}$$

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

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

Substitute $9.8m/s^2$for g,$36掳$ for $胃$, and$0.45$ for${渭}_k$ in the above equation.

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

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