Chapter 5, Problem 102A

(a)

To determine

The force exerted on the boulder to move it up the mountain with constant velocity.

Answer to Problem 102A

The force exerted on the boulder to move it up the mountain with constant velocity is $68\text{N}$ .

Explanation of Solution

Given:

The mass $\left(m\right)$ of the boulder is $20.0\text{kg}$ .

The coefficient of kinetic friction $\left({\mu }_{\text{k}}\right)$ is $0.40$ .

The angle made by the surface of the mountain with the horizontal is $\theta =30°$ .

Formula used:

The expression for the kinetic friction acting on a moving object is as follows:

  $F_{\text{f,kinetic}}={\mu }_{\text{k}}{F}_{\text{N}}$

Here, ${\mu }_{\text{k}}$ is the coefficient of kinetic friction and $F_{\text{N}}$ is the normal force acting on the object.

Calculation:

For the given specifications, the free body diagram of the situation is shown below.

  

Find the normal force $\left(F_{\text{N}}\right)$ acting on the boulder, by equating the forces acting in the $y$ direction,

  $\begin{array}{l}{F}_{\text{N}}=mg\cos \theta \\ F_{\text{N}}=\left(20.0\text{kg}\right)\times \left(9.80{\text{m/s}}^2\right)\times \cos \left(30°\right)\\ F_{\text{N}}=\left(20.0\text{kg}\right)\times \left(9.80{\text{m/s}}^2\right)\times \left(0.866\right)\\ F_{\text{N}}=170\text{kg}\cdot {\text{m/s}}^{\text{2}}\times \left(\frac{\text{1}\text{N}}{1\text{kg}\cdot {\text{m/s}}^{\text{2}}}\right)\\ F_{\text{N}}=170\text{N}\end{array}$

The kinetic frictional force acting on the person is,

  $\begin{array}{l}{f}_{\text{kinetic}}={\mu }_{\text{k}}{F}_{\text{N}}\\ f_{\text{kinetic}}=0.4\times 170\\ f_{\text{kinetic}}=68\text{N}\end{array}$

The person has to apply force equal to the kinetic friction to move the boulder up the hill.

Conclusion:

Therefore, the force exerted by the person S on the boulder to move it up the mountain with constant velocity is $68\text{N}$ .

(b)

To determine

The height of the mountain.

Answer to Problem 102A

The height of the mountain is $40.6\times {10}^8\text{m}$ .

Explanation of Solution

Given:

The mass $\left(m\right)$ of the boulder is

  $20.0\text{kg}$ .

The coefficient of kinetic friction w

  $\left({\mu }_{\text{k}}\right)$ is $0.40$ .

The angle made by the surface of the mountain with the horizontal is $\theta =30°$ .

The initial velocity of the person S is $u=0.25\text{m/s}$ .

The time taken by the person S to reach the top of the hill is $t=8\text{h}$ .

Formula used:

The distance travelled according to second equation of motion is,

  $y=ut+\frac{1}{2}a{t}^2$

Here, $u$ is the initial velocity of the object, $a$ is the acceleration and $t$ is the time taken by the object.

Calculation:

Consider the acceleration due to gravity of Earth is $g=9.80{\text{m/s}}^2$ .

Consider the height of the hill is denoted by $y$ .

The height of the hill as follows:

  $\begin{array}{l}y=\left(u\sin \theta \right)t+\frac{1}{2}g{t}^2\\ y=\left(0.25\text{m/s}\right)\sin \left(30°\right)\times \left(\text{8}\text{h}\times \frac{3600\text{s}}{1\text{h}}\right)+\frac{1}{2}\times \left(9.80{\text{m/s}}^2\right){\left(8\text{h}\times \frac{3600\text{s}}{1\text{h}}\right)}^2\\ y=3,600\text{m}+4,064,256,000\text{m}\\ y=40.6\times {10}^8\text{m}\end{array}$

Conclusion:

Therefore, the height of the hill is $40.6\times {10}^8\text{m}$ .