Chapter 5, Problem 40P

(a)

To determine

The friction force exerted by the road.

Answer to Problem 40P

The friction force exerted by the road is $\underset{\_}{3.95\times {10}^3\text{N}}$.

Explanation of Solution

Write the expression for drag force on the car.

    $R=\frac{1}{2}D\rho A{v}^2$        (I)

Here, $R$ is the dray force, $D$ is the drag coefficient, $\rho$ is the density, $A$ is the area, and $v$ is the speed of the car.

Frictional force and drag force are opposite in direction.

Using Newton’s second law, write the expression for net force in the horizontal direction.

    $f_{\text{s}}-R=ma$        (II)

Here, $f_{\text{s}}$ is the force due to static friction, $a$ is the acceleration in the horizontal direction, and $m$ is the mass.

Solve equation (III) for $f_{\text{s}}$.

    $f_{\text{s}}=R+ma$        (III)

Use equation (I) in (II).

    $f_{\text{s}}-\frac{1}{2}D\rho A{v}^2=ma$        (IV)

Conclusion:

Substitute, $0.34$ for $D$, $1.2\text{kg}/{\text{m}}^3$ for $\rho$, $2.6{\text{m}}^2$ for $A$, and $10\text{m}/\text{s}$ for $v$ in the equation (I), to find $R$.

    $\begin{array}{c}R=\frac{1}{2}\left(0.34\right)\left(1.2\text{kg}/{\text{m}}^3\right)\left(2.6{\text{m}}^2\right){\left(10\text{m}/\text{s}\right)}^2\\ =53.0\text{N}\end{array}$

Substitute, $53.0\text{N}$ for $R$, $1300\text{kg}$ for $m$, and $3\text{m}/{\text{s}}^2$ for $a$ in the equation (III), to find $f_{\text{s}}$.

    $\begin{array}{c}{f}_{\text{s}}=53.0\text{N}+\left(1300\text{kg}\right)\left(3\text{m}/{\text{s}}^2\right)\\ =3.95\times {10}^3\text{N}\end{array}$

Therefore, the friction force exerted by the road is $\underset{\_}{3.95\times {10}^3\text{N}}$.

(b)

To determine

The acceleration of the car if drag coefficient is $0.200$.

Answer to Problem 40P

The acceleration of the car if drag coefficient is $0.200$ is $\underset{\_}{3.02\text{m}/{\text{s}}^2}$.

Explanation of Solution

Conclusion:

Substitute, $3950\text{N}$ for $f_{\text{s}}$, $1300\text{kg}$ for $m$, $0.200$ for $D$, $1.2\text{kg}/{\text{m}}^3$ for $\rho$, $2.6{\text{m}}^2$ for $A$, and $10\text{m}/\text{s}$ for $v$ in the equation (IV), and solve for $a$.

    $\begin{array}{c}3950\text{N}-\frac{1}{2}\left(0.200\right)\left(1.2\text{kg}/{\text{m}}^3\right)\left(2.6{\text{m}}^2\right){\left(10\text{m}/\text{s}\right)}^2=\left(1300\text{kg}\right)a\\ a=\frac{3950\text{N}-31.2\text{N}}{\left(1300\text{kg}\right)}\\ =3.02\text{m}/{\text{s}}^2\end{array}$

Therefore, the acceleration of the car is $\underset{\_}{3.02\text{m}/{\text{s}}^2}$.

(c)

To determine

The possible maximum speed pf the car at $D=0.340$.

Answer to Problem 40P

The possible maximum speed pf the car at $D=0.340$ is $\underset{\_}{86.3\text{m}/\text{s}}$.

Explanation of Solution

At the maximum speed, the net force along the horizontal direction is zero.

Conclusion:

Substitute, $3950\text{N}$ for $f_{\text{s}}$, $0$ for $ma$, $0.34$ for $D$, $1.2\text{kg}/{\text{m}}^3$ for $\rho$, and, $2.6{\text{m}}^2$ for $A$ in the equation (IV), and solve for $v$.

    $\begin{array}{c}3950\text{N}-\frac{1}{2}\left(0.34\right)\left(1.2\text{kg}/{\text{m}}^3\right)\left(2.6{\text{m}}^2\right)v^2=0\\ v=\sqrt{\frac{3950}{0.530}\frac{\text{kg}\cdot \text{m}\cdot \text{m}}{{\text{s}}^2\cdot \text{kg}}}\\ =86.3\text{m}/\text{s}\end{array}$

Therefore, The possible maximum speed pf the car at $D=0.340$ is $\underset{\_}{86.3\text{m}/\text{s}}$.

(d)

To determine

The possible maximum speed pf the car at $D=0.200$.

Answer to Problem 40P

The possible maximum speed pf the car at $D=0.200$ is $\underset{\_}{113\text{m}/\text{s}}$.

Explanation of Solution

Conclusion:

Substitute, $3950\text{N}$ for $f_{\text{s}}$, $0$ for $ma$, $0.34$ for $D$, $1.2\text{kg}/{\text{m}}^3$ for $\rho$, and, $2.6{\text{m}}^2$ for $A$ in the equation (IV), and solve for $v$.

    $\begin{array}{c}3950\text{N}-\frac{1}{2}\left(0.200\right)\left(1.2\text{kg}/{\text{m}}^3\right)\left(2.6{\text{m}}^2\right)v^2=0\\ v=\sqrt{\frac{3950}{0.312}\frac{\text{kg}\cdot \text{m}\cdot \text{m}}{{\text{s}}^2\cdot \text{kg}}}\\ =113\text{m}/\text{s}\end{array}$

The drag reduction significantly affects maximum speed.

Therefore, the possible maximum speed pf the car at $D=0.200$ is $\underset{\_}{113\text{m}/\text{s}}$.