Chapter 2.8, Problem 36P

(a)

To determine

The total power required to accelerate the car on the uphill road at a constant velocity.

Answer to Problem 36P

The total power required to accelerate the car on the uphill road at a constant velocity is $\text{47}\text{kW}$.

Explanation of Solution

The total work required to accelerate the car on the uphill road is equal to the summation of the rates of changes in kinetic and potential energies respectively.

Write the formula to calculate rate of change in kinetic energy of the car $\left({\dot{W}}_a\right)$.

${\dot{W}}_a=\frac{1}{2}\frac{m\left(V_2^2-V_1^2\right)}{\Delta t}$ (I)

Here, mass of the car is $m$, initial and final velocities of the car are $V_1$ and $V_2$ respectively and change in time interval is $\Delta t$.

Write the formula to calculate rate of change in potential energy of the car $\left({\dot{W}}_g\right)$.

${\dot{W}}_g=\frac{mg\Delta z}{\Delta t}$ (II)

Here, acceleration due to gravity is $g$ and change in vertical elevation of the car is $\Delta z$.

Write the formula to calculate total power required to accelerate the car on the uphill road $\left({\dot{W}}_{\text{total}}\right)$.

${\dot{W}}_{\text{total}}={\dot{W}}_a+{\dot{W}}_g$ (III)

Conclusion:

Since the velocity is constant, the initial and final velocities of the car will be zero.

$V=V_1=V_2=0$

Then the rate of change in kinetic energy of the car will also be zero.

${\dot{W}}_a=0\text{kW}$

Substitute $\text{1150}\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^2$ for $g$, $50\text{m}$ for $\Delta z$ and $12\text{s}$ for $\Delta t$ in Equation (II).

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

Substitute $0\text{kW}$ for ${\dot{W}}_a$ and $47\text{kW}$ for ${\dot{W}}_g$ in Equation (III).

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

The total power required to accelerate the car on the uphill road at a constant velocity is $\text{47}\text{kW}$.

(b)

To determine

The total power required to accelerate the car on the uphill road from rest to the final velocity.

Answer to Problem 36P

The total power required to accelerate the car on the uphill road from rest to the final velocity is $\text{90}\text{.1}\text{kW}$.

Explanation of Solution

Conclusion:

Substitute $\text{1150}\text{kg}$ for $m$, $30\text{m}/\text{s}$ for $V_2$, $0\text{m}/\text{s}$ for $V_1$ and $12\text{s}$ for $\Delta t$ in Equation (I).

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

Substitute $43.1\text{kW}$ for ${\dot{W}}_a$ and $47\text{kW}$ for ${\dot{W}}_g$ in Equation (III).

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

The total power required to accelerate the car on the uphill road from rest to the final velocity is $\text{90}\text{.1}\text{kW}$.

(c)

To determine

The total power required to accelerate the car on the uphill road from initial to the final velocity.

Answer to Problem 36P

The total power required to accelerate the car on the uphill road from initial to the final velocity is $-\text{10}\text{.5}\text{kW}$.

Explanation of Solution

Conclusion:

Substitute $\text{1150}\text{kg}$ for $m$, $5\text{m}/\text{s}$ for $V_2$, $35\text{m}/\text{s}$ for $V_1$ and $12\text{s}$ for $\Delta t$ in Equation (I).

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

Substitute $-57.5\text{kW}$ for ${\dot{W}}_a$ and $47\text{kW}$ for ${\dot{W}}_g$ in Equation (III).

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

The total power required to accelerate the car on the uphill road from initial to the final velocity is $-\text{10}\text{.5}\text{kW}$.