91Ó°ÊÓ

Mass of the rock$=520kg$

Hillside is $500m$long and $300m$high

Coefficient of kinetic friction between the rock and the hill surface is 0.25.

Potential energy, $\mathit{U}\mathbf{=}\mathit{m}\mathit{g}\mathit{h}$and kinetic energy, $\mathit{k}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$. Thermal energy transferred will be equal to the work done against friction.

The initial potential energy is

$$\begin{gathered} U_i=mg{y}_i \\ =\left(520\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(300\mathrm{m}\right) \\ =1.53×{10}^6?J \\ =1.53\mathrm{MJ} \end{gathered}$$

where + y is upward and y = 0at the bottom (so that $U_f=0$).

Hence, the initial potential energy is 1.53 MJ

Since $f_k={\mathrm{μ}}_k{F}_N={\mathrm{μ}}_{k}mg\cos \mathrm{θ}$

we have

role="math" localid="1661229424523" $∆E_{th}=f_{k}d={\mathrm{μ}}_{k}mgd\cos \mathrm{θ}$from Eq. 8-31.

Now, the hillside surface (of length d = 500 m) is treated as an hypotenuse of a 3-4-5 triangle, so $\cos \mathrm{θ}=\mathrm{x}/\mathrm{d}\mathrm{where}\mathrm{x}=400\mathrm{m}$. Therefore,

$$\begin{gathered} ∆E_{th}={\mathrm{μ}}_{k}mgd\frac{x}{d} \\ ={\mathrm{μ}}_{k}mgx \\ =\left(0.25\right)\left(520\right)\left(9.8\right)\left(400\right) \\ =5.1×{10}^5\mathrm{J} \\ =0.51\mathrm{MJ} \end{gathered}$$

Hence the energy is 0.51 MJ

Using Eq. 8-31 (with W = 0) we find

$$\begin{gathered} K_f=K_i+U_i-U_f-∆E_{th} \\ =0+\left(1.53×{10}^6\mathrm{J}\right)-0-\left(1.53×{10}^6\mathrm{J}\right) \\ =1.02×{10}^6\mathrm{J} \\ =1.02\mathrm{MJ} \end{gathered}$$

Hence the kinetic energy is 1.0 MJ

From $$\begin{gathered} K_f=\frac{1}{2}m{v}^2, \\ \mathrm{Where}{\mathrm{K}}_{\mathrm{f}}=1.02×{10}^6\mathrm{J}\mathrm{and}m=520\mathrm{kg} \end{gathered}$$

we obtain v = 63 m/s

Hence the speed is 63 m/s