Chapter 3, Problem 35P

a)

To determine

The power required for the car to climb the uphill road at the constant velocity.

Explanation of Solution

Given:

Mass of the car $\left(m\right)$ is $1150\text{kg}$.

Height climbed by the car in the uphill road $\left(h\right)$ is $\text{100}\text{m}$.

Slope of the road $\left(\theta \right)$ is $30°$.

Time interval $\left(\Delta t\right)$ is $12\text{s}$.

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

Calculation:

Calculate the vertical rise of the car $\left(\Delta z\right)$.

  $\begin{array}{c}\Delta z=h\sin \theta \\ =\left(\text{100}\text{m}\right)\left(\sin 30°\right)\\ =\text{50}\text{m}\end{array}$

The change in kinetic energy of the car is zero because the velocity of the car is constant.

  ${\dot{W}}_a=0\text{kW}$

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

  $\begin{array}{c}{\dot{W}}_g=mg\frac{\Delta z}{\Delta t}\\ =\left(1150\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(\frac{50\text{m}}{12\text{s}}\right)\\ =\left(1150\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(\frac{50\text{m}}{12\text{s}}\right)\left(\frac{1\text{kJ}}{1000\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =47\text{kW}\end{array}$

Calculate the power required for the car to climb the uphill road at the constant velocity $\left({\dot{W}}_{\text{total}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{total}}={\dot{W}}_a+{\dot{W}}_g\\ =0\text{kW}+47\text{kW}\\ =47\text{kW}\end{array}$

Thus, the power required for the car to climb the uphill road at the constant velocity is $\underset{\_}{\text{47}\text{kW}}$.

b)

To determine

The power required for the car to climb the uphill road from rest to the final velocity.

Explanation of Solution

Given:

Initial velocity of the car $\left(V_1\right)$ is $0\text{m}/\text{s}$.

Final velocity of the car $\left(V_2\right)$ is $30\text{m}/\text{s}$.

Calculation:

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

  $\begin{array}{c}{\dot{W}}_a=\frac{1}{2}m\frac{\left(V_2^2-V_1^2\right)}{\Delta t}\\ =\frac{1}{2}\left(1150\text{kg}\right)\frac{\left[{\left(30\text{m}/\text{s}\right)}^2-{\left(0\text{m}/\text{s}\right)}^2\right]}{12\text{s}}\\ =\frac{1}{2}\left(1150\text{kg}\right)\frac{\left[{\left(30\text{m}/\text{s}\right)}^2-{\left(0\text{m}/\text{s}\right)}^2\right]}{12\text{s}}\left(\frac{1\text{kJ}}{1000\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =43.1\text{kW}\end{array}$

Calculate the power required for the car to climb the uphill road rest to the final velocity $\left({\dot{W}}_{\text{total}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{total}}={\dot{W}}_a+{\dot{W}}_g\\ =43.1\text{kW}+47\text{kW}\\ =90.1\text{kW}\end{array}$

Thus, the power required for the car to climb the uphill road from rest to the final velocity is $\underset{\_}{90.1\text{kW}}$.

c)

To determine

The power required for the car to climb the uphill road from initial velocity to the final velocity.

Explanation of Solution

Given:

Initial velocity of the car $\left(V_1\right)$ is $35\text{m}/\text{s}$.

Final velocity of the car $\left(V_2\right)$ is $5\text{m}/\text{s}$.

Calculation:

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

  $\begin{array}{c}{\dot{W}}_a=\frac{1}{2}m\frac{\left(V_2^2-V_1^2\right)}{\Delta t}\\ =\frac{1}{2}\left(1150\text{kg}\right)\frac{\left[{\left(5\text{m}/\text{s}\right)}^2-{\left(35\text{m}/\text{s}\right)}^2\right]}{12\text{s}}\\ =\frac{1}{2}\left(1150\text{kg}\right)\frac{\left[{\left(5\text{m}/\text{s}\right)}^2-{\left(35\text{m}/\text{s}\right)}^2\right]}{12\text{s}}\left(\frac{1\text{kJ}}{1000\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =-57.5\text{kW}\end{array}$

Calculate the power required for the car to climb the uphill road from initial velocity to the final velocity $\left({\dot{W}}_{\text{total}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{total}}={\dot{W}}_a+{\dot{W}}_g\\ =-57.5\text{kW}+47\text{kW}\\ =-10.5\text{kW}\end{array}$

Thus, the power required for the car to climb the uphill road from initial velocity to the final velocity is $\underset{\_}{-10.5\text{kW}}$.