Chapter 11.2, Problem 18PP

(a)

To determine

The velocity of the skier at the bottom of the valley.

Answer to Problem 18PP

  $v_{bv}=29.69\text{ m/s}$

Explanation of Solution

Given:

The velocity of a skier at the top of the first hill is $v_{t1}=0\text{ m/s}$

The height of the first hill from which a skier starts to move is $h_1=45\text{ m}$

A skier skis down a $\text{30}°$ incline in to a valley.

Formula used:

According to law of conservation of energy, the total energy is always conserved. This means that the system energy will always remain same.

Calculation:

Since the energy is conserved, the sum of kinetic energy and potential energy of a skier at the top of the first hill is equal to the sum of kinetic energy and potential energy of that skier at the bottom of the valley. This can written as,

  $K{E}_{t1}+P{E}_{t1}=K{E}_{bv}+P{E}_{bv}$  $\left(1\right)$

Where,

$K{E}_{t1}$ is the kinetic energy of a skier at the top of the first hill, $K{E}_{t1}=\frac{1}{2}m{v}_{t1}^2$

$P{E}_{t1}$ is the potential energy of a skier at the top of the first hill, $P{E}_{t1}=mg{h}_1$

$K{E}_{bv}$ is the kinetic energy of a skier at the bottom of the valley, $K{E}_{bv}=\frac{1}{2}m{v}_{bv}^2$

$P{E}_{bv}$ is the potential energy of a skier at the bottom of the valley, $P{E}_{bv}=mg{h}_{bv}$

Substituting these expressions in equation $\left(1\right)$ ,

  $\frac{1}{2}m{v}_{t1}^2+mg{h}_1=\frac{1}{2}m{v}_{bv}^2+mg{h}_{bv}$  $\left(2\right)$

$m$ is the mass of a skier

g is the acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

$v_{t1}$ and $v_{bv}$ are the velocity of a skier at the top of the first hill and bottom of the valley respectively.

Since the velocity of a skier at the bottom of the valley is to be determined, potential energy, $P{E}_{bv}$ is zero (because the height is zero). Also $v_{t1}=0\text{ m/s}$ . Therefore equation $\left(2\right)$ becomes,

  $0+mg{h}_1=\frac{1}{2}m{v}_{bv}^2+0$

  $\Rightarrow v_{bv}^2=2g{h}_1$

  $\Rightarrow v_{bv}=\sqrt{2g{h}_1}$  $\left(3\right)$

Substituting the numerical values in equation (3),

  $\begin{array}{c}{v}_{bv}=\sqrt{2\left(9.8{\text{ m/s}}^2\right)\left(45.0\text{ m}\right)}\\ =\sqrt{882{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =29.69\text{m}/\text{s}\end{array}$

Conclusion:

The velocity of the skier at the bottom of the valley is $29.69\text{ m/s}$ .

(b)

To determine

The skier’s speed at the top of the second hill.

Answer to Problem 18PP

  $v_{t2}=9.89\text{ m/s}$

Explanation of Solution

Given:

The velocity of a skier at the top of the first hill is $v_{t1}=0\text{ m/s}$

The height of the first hill from which a skier starts to move is $h_1=45\text{ m}$

A skier skis down a $\text{30}°$ incline in to a valleyand he continues up a second hill.

Height of the second hill is $h_2=40.0\text{ m}$

Formula used:

According to law of conservation of energy, the total energy is always conserved. This means that the system energy will always remain same.

Calculation:

Energy equation for a skier moving from the top of the first hill to the top of the second hill can be written as,

  $K{E}_{t1}+P{E}_{t1}=K{E}_{t2}+P{E}_{t2}$  $\left(4\right)$

Where,

$K{E}_{t1}$ is the kinetic energy of a skier at the top of the first hill, $K{E}_{t1}=\frac{1}{2}m{v}_{t1}^2$

$P{E}_{t1}$ is the potential energy of a skier at the top of the first hill, $P{E}_{t1}=mg{h}_1$

$K{E}_{t2}$ is the kinetic energy of a skier at the bottom of the valley, $K{E}_{t2}=\frac{1}{2}m{v}_{t2}^2$

$P{E}_{t2}$ is the potential energy of a skier at the bottom of the valley, $P{E}_{t2}=mg{h}_2$

Substituting these expressions in equation $\left(1\right)$ ,

  $\frac{1}{2}m{v}_{t1}^2+mg{h}_1=\frac{1}{2}m{v}_{t2}^2+mg{h}_2$  $\left(5\right)$

$m$ is the mass of a skier

$g$ is the acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

$v_{t1}$ and $v_{t2}$ are the velocity of a skier at the top of the first hill and at the top of the second hill respectively.

Since $v_{t1}=0\text{ m/s}$ , equation $\left(5\right)$ becomes,

  $0+mg{h}_1=\frac{1}{2}m{v}_{t2}^2+mg{h}_2$

  $\Rightarrow \frac{1}{2}m{v}_{t2}^2=mg{h}_1-mg{h}_2$

  $\Rightarrow \frac{1}{2}{v}_{t2}^2=g{h}_1-g{h}_2$

  $\Rightarrow v_{t2}^2=2g\left(h_1-h_2\right)$

  $\Rightarrow v_{t2}=\sqrt{2g\left(h_1-h_2\right)}$  $\left(6\right)$

Substituting the numerical values in equation $\left(6\right)$ ,

  $\begin{array}{c}{v}_{t2}=\sqrt{2\left(9.8{\text{ m/s}}^2\right)\left(45.0\text{ m}-40.0\text{ m}\right)}\\ =\sqrt{98{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =9.89\text{m}/\text{s}\end{array}$

Conclusion:

The skier’s speed at the top of the second hill is $9.89\text{ m/s}$ .

(c)

To determine

To identify: Whether the angles of the hills affect the values obtained in part (a) and (b).

Answer to Problem 18PP

No, the angles of the hills will not affect the values obtained in part (a) and (b)

Explanation of Solution

As can be seen in the equations used to calculate speed of the skier at the bottom of the valley and top of the second hill, the gravitational potential energy of a skier at the top of the hill depends merely on the height of the hill. It does not depend on the angle of the hill or length of the pathway along the hill. Therefore the angles of the hills will not affect the velocity of a skier at the bottom of the valley and at the top of the second hill.