Chapter 6, Problem 95P

(a)

To determine

The power must the engine deliver to drive the car on a level road at $20.0\text{m}/\text{s}$.

Answer to Problem 95P

The power must the engine deliver to drive the car on a level road at $20.0\text{m}/\text{s}$ is $10\text{kW}$.

Explanation of Solution

Take $+x$ be up the incline.

Write the expression for the net force along $x$ direction.

$\Sigma F_x=F_{\text{air}}-mg\sin \varphi =0$

Here $\Sigma F_x$ is the net force along $x$ direction, $F_{\text{air}}$ is the air resistance, $m$ is the mass of car, $g$ is the acceleration due to gravity and $\varphi$ is the inclination of hill road.

Rearrange the above equation to get $F_{\text{air}}$.

$F_{\text{air}}=mg\sin \varphi$

Write the expression for rate at which air resistance dissipates energy.

$\begin{array}{c}{P}_{\text{air}}=F_{\text{air}}v\cos 180°\\ =-Fv\end{array}$

Here, $P_{\text{air}}$ is the rate at which air resistance dissipates energy and $v$ is the velocity of car .

The $\cos 180°$ is introduced since force of air resistance is opposite the car’s velocity.

Conclusion:

Substitute $mg\sin \varphi$ for $F_{\text{air}}$ in above equation to get $P_{\text{air}}$.

$P_{\text{air}}=-mg\sin \varphi v$

The car has to give power equivalent to $P_{\text{air}}$ in opposite direction to drive the car in constant velocity.

Therefore, write the expression for the power the engine must deliver to drive the car on level ground.

$\begin{array}{c}{P}_{\text{engine}}=-P_{\text{air}}\\ =-mgv\sin \varphi \end{array}$

Substitute $1500\text{kg}$ for $m$ , $9.80\text{m}/{\text{s}}^2$ for $g$ , $20.0\text{m}/\text{s}$ for $v$ and $2.0°$ for $\theta$ in above equation to get $P_{\text{engine}}$.

$\begin{array}{c}{P}_{\text{engine}}=\left(1500\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(20.0\text{m}/\text{s}\right)\sin 2.0°\\ =10260\text{W}\times \frac{1\text{kW}}{1000\text{W}}\\ =10\text{kW}\end{array}$

Therefore, the power must the engine deliver to drive the car on a level road at $20.0\text{m}/\text{s}$ is $10\text{kW}$.

(b)

To determine

The steepest hill the car can climb at $20.0\text{m}/\text{s}$, if maximum power that can delivered by the engine is $40.0\text{kW}$.

Answer to Problem 95P

The car can climb at $20.0\text{m}/\text{s}$ if steep of the hill is $5.8°$.

Explanation of Solution

Write the expression power available to climb the hill.

$P=P_{\text{max}}-P_{\text{air}}$ (I)

Here, $P$ is the power available to climb the hill, $P_{\text{max}}$ is the maximum power that engine can deliver and $P_{\text{air}}$ is the rate t which air resistance dissipates the energy.

Write the expression for the power delivered by the engine.

$P=mgv\sin \varphi$ (II)

Conclusion:

Substitute $40.0\text{kW}$ for $P_{\text{max}}$ and $10.26\text{kW}$ for $P_{\text{air}}$ in equation(I) to get $P$.

$\begin{array}{c}P=40.0\text{kW}-10.26\text{kW}\\ =\text{29}\text{.74}\text{kW}\end{array}$

Substitute $1500\text{kg}$ for $m$ , $9.80\text{m}/{\text{s}}^2$ for $g$ , $20.0\text{m}/\text{s}$ for $v$ and $29.74\text{kW}$ for $P$ in above equation to get $P$.

$\begin{array}{c}29.74\text{kW}=\left(1500\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(20.0\text{m}/\text{s}\right)\sin \varphi \\ \sin \varphi =\frac{29.74\text{kW}}{\left(1500\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(20.0\text{m}/\text{s}\right)}\\ =5.8°\end{array}$

Therefore, the car can climb at $20.0\text{m}/\text{s}$ if steep of the hill is $5.8°$.