Chapter 7, Problem 109P

To determine

The speed of the boy at the bottom.

Answer to Problem 109P

The speed of the boy at the bottom is $\underset{\_}{13\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for the conservation of energy of the boy alone.

$K_i+U_i+W_{nc}=K_f+U_f$ (I)

Here, $K_i$ is the initial kinetic energy, $U_i$ is the initial potential energy, $W_{nc}$ is the work done before the second friend jumps to the sled, $K_f$ is the final kinetic energy and $U_f$ is the final potential energy

Write the expression for the initial kinetic energy.

$K_i=0$

Write the expression for the initial potential energy.

$\begin{array}{c}{U}_i=m_{1}g{h}_1\\ =m_{1}g{d}_1\text{sin}\theta \end{array}$

Here, $m_1$ is the mass of the boy, $g$ is the acceleration due to gravity and $h_1$ is the initial height of the boy along the slope.

Write the expression to find the work done before the second friend jumps to the sled.

$W_{nc}=\mu m_{1}g{d}_1\text{cos}\theta$

Write the expression for the final kinetic energy.

$K_f=\frac{1}{2}{m}_1{v}_1^2$

Write the expression for the final potential energy.

$U_f=0$

Rewrite equation (I) to find the velocity of the boy.

$\begin{array}{c}0+m_{1}g{d}_1\text{sin}\theta +\mu m_{1}g{d}_1\text{cos}\theta =\frac{1}{2}{m}_1{v}_1^2+0\\ v_1=\sqrt{2g{d}_1\left(\text{sin}\theta -\mu \text{cos}\theta \right)}\end{array}$ (II)

Write the expression for conservation of momentum and rewrite it to find the initial velocity of both the boys.

$\begin{array}{c}{m}_1{v}_1=\left(m_1+m_2\right)v_2\\ v_2=\frac{m_1{v}_1}{\left(m_1+m_2\right)}\end{array}$ (III)

Write the expression for the conservation of energy of the two boys.

$K_i+U_i+W_{friction}=K_f+U_f$ (IV)

Write the expression for the initial kinetic energy of two boys.

$K_i=\frac{1}{2}\left(m_1+m_2\right)v_2^2$

Write the expression for the initial potential energy of two boys.

$U_i=\left(m_1+m_2\right)g{h}_2$

Write the expression to find the work done after the second friend jumps to the sled.

$W_{friction}=\mu \left(m_1+m_2\right)g{d}_2\text{cos}\theta$

Write the expression for the final kinetic energy.

$K_f=\frac{1}{2}\left(m_1+m_2\right)v_3^2$

Here, $v_3$ is the final velocity.

Write the expression for the final potential energy.

$U_f=0$

Rewrite equation (IV) to find the final velocity of the boys.

$\begin{array}{c}\begin{array}{l}\frac{1}{2}\left(m_1+m_2\right)v_2^2+\left(m_1+m_2\right)g{h}_2\\ +\mu \left(m_1+m_2\right)g{d}_2\text{cos}\theta \end{array}\}=\frac{1}{2}\left(m_1+m_2\right)v_3^2+0\\ v_3=\sqrt{2g{d}_2\left(\text{sin}\theta -\mu \cos \theta \right)+v_2^2}\end{array}$ (V)

Conclusion:

Substitute $9.8\text{m}/s^2$ for $g$, $20\text{m}$ for $d_1$, $0.12$ for $\mu$ and $15°$ for $\theta$ in equation (II).

$\begin{array}{c}{v}_1=\sqrt{2\left(9.8\text{m}/s^2\right)\left(20\text{m}\right)\left(\text{sin}15°-\left(0.12\right)\text{cos}15°\right)}\\ =7.48\text{m}/\text{s}\end{array}$

Substitute $7.48\text{m}/\text{s}$ for $v_1$, $60\text{kg}$ for $m_1$ and $50\text{kg}$ for $m_2$ in equation (III).

$\begin{array}{c}{v}_2=\frac{\left(60\text{kg}\right)\left(7.48\text{m}/\text{s}\right)}{\left(60\text{kg}+50\text{kg}\right)}\\ =4.08\text{m}/\text{s}\end{array}$

Substitute $9.8\text{m}/s^2$ for $g$, $50\text{m}$ for $d_2$, $4.08\text{m}/\text{s}$ for $v_2$, $0.12$ for $\mu$ and $15°$ for $\theta$ in equation (V).

$\begin{array}{c}{v}_3=\sqrt{2\left(9.8\text{m}/s^2\right)\left(50\text{m}\right)\left(\text{sin}15°-\left(0.12\right)\cos 15°\right)+{\left(4.08\text{m}/\text{s}\right)}^2}\\ =13\text{m}/\text{s}\end{array}$

Therefore, the speed of the boy at the bottom is $\underset{\_}{13\text{m}/\text{s}}$.