Chapter 2.8, Problem 34P

A damaged 1200-kg car is being towed by a truck. Neglecting the friction, air drag, and rolling resistance, determine the extra power required (a) for constant velocity on a level road, (b) for constant velocity of 50 km/h on a 30° (from horizontal) uphill road, and (c) to accelerate on a level road from stop to 90 km/h in 12 s.

To determine

The extra power required for constant velocity on a level road.

Answer to Problem 34P

The extra power required for constant velocity on a level road is $\underset{\_}{0}$.

Explanation of Solution

Write the equation of total power required during the rates of changes in potential and kinetic energies.

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

Here, the power needed to acceleate the body is ${\dot{W}}_a$ and power need to raise the mass of damaged 1200 kg car by a truck is ${\dot{W}}_g$.

Conclusion:

Since the velocity is constant on a level road, the extra power required will be considered as zero.

${\dot{W}}_{\text{total}}=0$

Thus, the extra power required for constant velocity on a level road is $\underset{\_}{0}$.

To determine

The extra power required for constant velocity of 50 km/h on a $30°$ uphill road.

Answer to Problem 34P

The extra power required for constant velocity of 50 km/h on a $30°$ uphill road is $\underset{\_}{\text{81}\text{.7}\text{kW}}$.

Explanation of Solution

Calculate the value of ${\dot{W}}_g$.

$\begin{array}{c}{\dot{W}}_g=\frac{mg\left(z_2-z_1\right)}{t_2-t_1}\\ =\frac{mg\left(\Delta z\right)}{\Delta t}\\ =mg{V}_z\\ =mgV\sin \theta \end{array}$ (II)

Here, mass of a damaged car is m, acceleration due to gravity is g, difference between the elevation of ski lift is $z_2-z_1$, angle of uphill road is $\theta$, velocity in z axes is $V_z$, initial and final time is $t_1$ and $t_2$ respectively.

Conclusion:

Since the velocity is constant on a level road, the extra power needed to acceleate the body is considered as zero.

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

Substitute 1200 kg for m, $9.81{\text{m/s}}^{\text{2}}$ for g, 50 km/h for V, and $30°$ for $\theta$ in Equation (II).

$\begin{array}{c}{\dot{W}}_g=\left(\text{1200}\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\left(\text{50}\frac{\text{km}}{\text{h}}\right)\sin 30°\\ =\left(\text{1200}\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\left(\text{50}\frac{\text{km}\times \text{1000}\text{m}}{\text{h}\times \text{3600}\frac{\text{s}}{1\text{h}}}\right)0.5\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =81.7\text{kW}\end{array}$

Substitute 0 for ${\dot{W}}_a$ and 81.7 kW for ${\dot{W}}_g$ in Equation (I).

$\begin{array}{c}{\dot{W}}_{\text{total}}=81.7\text{kW}+0\\ =81.7\text{kW}\end{array}$

Thus, the extra power required for constant velocity of 50 km/h on a $30°$ uphill road is $\underset{\_}{\text{81}\text{.7}\text{kW}}$.

To determine

The extra power required to accelerate on a level road from stop to 90 km/h in 12 s.

Answer to Problem 34P

The extra power required to accelerate on a level road from stop to 90 km/h in 12 s is $\underset{\_}{31.3\text{kW}}$.

Explanation of Solution

Calculate the value of ${\dot{W}}_a$.

${\dot{W}}_a=\frac{1}{2}\frac{m\left(V_2^2-V_1^2\right)}{t_2-t_1}$ (III)

Here, initial and final velocity of a car is $V_1$ and $V_2$ respectively.

Conclusion:

Since the car is accelerated on a level road, the value of ${\dot{W}}_g$ is considered to be zero.

${\dot{W}}_g=0$

Substitute 1200 kg for m, 12 s for $t_2$, 0 for $t_1$, 90 km/h for $V_2$, and 0 for $V_1$ in Equation (III).

$\begin{array}{c}{\dot{W}}_a=\frac{1}{2}\frac{\left(1200\text{kg}\right)\left({\left(90\text{km/h}\right)}^2-0\right)}{12\text{s}-0}\\ =600\text{kg}\left(\frac{{\left(90\frac{\text{km}\times \text{1000}\frac{\text{m}}{1\text{km}}}{\text{h}\times \text{3600}\frac{\text{s}}{1\text{h}}}\right)}^2}{12\text{s}}\right)\left(\frac{\text{1}\text{kJ/kg}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =31.3\text{kW}\end{array}$

Substitute 0 for ${\dot{W}}_g$ and 31.3 kW for ${\dot{W}}_a$ in Equation (I).

$\begin{array}{c}{\dot{W}}_{\text{total}}=31.3\text{kW}+0\\ =31.3\text{kW}\end{array}$

Thus, the extra power required to accelerate on a level road from stop to 90 km/h in 12 s is $\underset{\_}{31.3\text{kW}}$.