Chapter 3, Problem 33P

To determine

The power required to operate the ski lift.

The power required to accelerate the ski lift in $5\text{s}$ to its operating speed.

Explanation of Solution

Given:

Mass of each chair $\left(m\right)$ is $\text{250}\text{kg}$.

Elevation difference of the ski lift $\left(\Delta z\right)$ is $200\text{m}$.

Operating speed of the lift $\left(V\right)$ is $10\text{km}/\text{h}$.

One-way length of the ski lift $\left(D\right)$ is $\text{1}\text{km}$.

Acceleration due to gravity $\left(g\right)$ is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the number of chairs at any given time $\left(N\right)$.

  $\begin{array}{c}N=\frac{1\text{km}}{20\text{m}}\\ =\frac{1\text{km}\times \text{1000}\left(\frac{\text{m}}{1\text{km}}\right)}{20\text{m}}\\ =50\text{chairs}\end{array}$

Calculate the load of the lift at any given time $\left(m_{\text{load}}\right)$.

  $\begin{array}{c}{m}_{\text{load}}=N\left(m\right)\\ =\left(\text{50}\text{chairs}\right)\left(250\text{kg}/\text{chair}\right)\\ =12,500\text{kg}\end{array}$

Calculate the work required to raise the total load $\left(W_g\right)$.

  $\begin{array}{c}{W}_g=m_{\text{load}}g\left(z_2-z_1\right)\\ =\left(12,500\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(200\text{m}\right)\\ =\left(12,500\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(200\text{m}\right)\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^{\text{2}}}\right)\\ =24,525\text{kJ}\end{array}$

Calculate the change in time $\left(\Delta t\right)$.

  $\begin{array}{c}\Delta t=\frac{D}{V}\\ =\frac{1\text{km}}{10\text{km}/\text{h}}\\ =\left(0.1\text{h}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)\\ =360\text{s}\end{array}$

Calculate the power required to operate the ski lift $\left({\dot{W}}_g\right)$.

  $\begin{array}{c}{\dot{W}}_g=\frac{W_g}{\Delta t}\\ =\frac{24,525\text{kJ}}{360\text{s}}\\ =68.1\text{kW}\end{array}$

Thus, the power required to operate the ski lift is $\underset{\_}{\text{68}\text{.1}\text{kW}}$.

It is asked to determine the power required to accelerate the ski lift after $5\text{s}$.

  $t_2=5\text{s}$

Calculate the acceleration of the ski lift $\left(a\right)$.

  $a=\frac{\Delta V}{\Delta t}$

  $\begin{array}{c}a=\frac{V_2-V_1}{t_2-t_1}\\ =\frac{10\text{km}/\text{h}-0\text{km}/\text{h}}{5\text{s}-0\text{s}}\\ =\frac{\left(10\times \frac{5}{18}\text{m}/\text{s}\right)}{5\text{s}}\\ =0.556{\text{m}}^2/\text{s}\end{array}$

Calculate the rate of change in kinetic energy of the ski lift $\left({\dot{W}}_a\right)$.

  $\begin{array}{c}{\dot{W}}_a=\frac{1}{2}\frac{m\left(V_2^2-V_1^2\right)}{t_2-t_1}\\ =\frac{1}{2}\frac{\left(12,500\text{kg}\right)\left[{\left(2.\text{778}\text{m}/\text{s}\right)}^2-\left(0\text{m}/\text{s}\right)\right]}{5\text{s}-0}\\ =\frac{1}{2}\frac{\left(12,500\text{kg}\right){\left(2.\text{778}\text{m/s}\right)}^2}{5\text{s}}\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =9.6\text{kW}\end{array}$

Calculate the vertical distance travelled during the acceleration $\left(h\right)$.

  $\begin{array}{c}h=\frac{1}{2}a{t}^2\sin \alpha \\ =\frac{1}{2}\left(0.556\text{m}/{\text{s}}^2\right){\left(5\text{s}\right)}^2\left(\frac{200\text{m}}{1\text{km}}\right)\\ =\frac{1}{2}\left(0.556\text{m}/{\text{s}}^2\right){\left(5\text{s}\right)}^2\left(\frac{200\text{m}}{1000\text{m}}\right)\\ =1.39\text{m}\end{array}$

Calculate the rate of change in potential energy of the ski lift $\left({\dot{W}}_g\right)$.

  $\begin{array}{c}{\dot{W}}_g=\frac{mg\left(h\right)}{t_2-t_1}\\ =\frac{\left(\text{12,500}\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(1.39\text{m}\right)}{5\text{s}-0}\\ =\frac{\left(\text{12,500}\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(1.39\text{m}\right)}{5\text{s}}\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =\text{34}\text{.1}\text{kW}\end{array}$

Calculate the total power required to accelerate the ski lift $\left({\dot{W}}_{\text{total}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{total}}={\dot{W}}_a+{\dot{W}}_g\\ =9.6\text{kW}+34.1\text{kW}\\ =43.7\text{kW}\end{array}$

Thus, the power required to accelerate the ski lift in $5\text{s}$ to its operating speed is $\underset{\_}{43.1\text{kW}}$.