Question

During a rockslide, a 520 kgrock slides from rest down a hillside that islong and 300 mhigh. The coefficient of kinetic friction between the rock and the hill surface is 0.25. (a) If the gravitational potential energy Uof the rock–Earth system is zero at the bottom of the hill, what is the value of U just before the slide? (b) How much energy is transferred to thermal energy during the slide? (c) What is the kinetic energy of the rock as it reaches the bottom of the hill? (d) What is its speed then?

Short answer

$$\begin{gathered} \mathrm{a})U_i=1.53\mathrm{MJ} \\ \mathrm{b})∆{\mathrm{E}}_{\mathrm{th}}=0.51\mathrm{MJ} \\ \mathrm{c}){\mathrm{K}}_{\mathrm{f}}=1.0\mathrm{MJ} \\ \mathrm{d})\mathrm{v}=63\mathrm{m}/\mathrm{s} \end{gathered}$$

Step-by-step solution

Given Data

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.

Understanding the concept

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.

Step 3(a): Calculate the value of U just before the slide

The initial potential energy is

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

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

Hence, the initial potential energy is 1.53 MJ

Step 4(b): Calculate how much energy is transferred to thermal energy during the slide 

Since $f_k={\mu }_k{F}_N={\mu }_{k}mg\cos \mathrm{\theta }$

we have

role="math" localid="1661229424523" $∆E_{th}=f_{k}d={\mu }_{k}mgd\cos \mathrm{\theta }$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 $\cos \mathrm{\theta }=\mathrm{x}/\mathrm{d}\mathrm{where}\mathrm{x}=400\mathrm{m}$. Therefore,

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

Hence the energy is 0.51 MJ

Step 5(c): Calculate the kinetic energy of the rock as it reaches the bottom of the hill

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

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

Hence the kinetic energy is 1.0 MJ

Step 6(d): Calculate the speed

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

we obtain v = 63 m/s

Hence the speed is 63 m/s