Chapter 6, Problem 55P

(a)

To determine

The value of the constant C.

Explanation of Solution

Given:

The mass of the car is $1220\text{Kg}$ . The constant velocity of the car is $55\text{mi}/\text{h}$ . Power of the car while cruising is $13.5\text{hp}$ . The coefficient of rolling friction is 0.0150.

Formula used:

Write the expression for the power of an object.

  $P=Fv$

Here, $P$ is the power, $F$ is the force applied on the object and $v$ is the velocity of the particle.

Draw the diagram to show the different forces acting on the car.

  

Write the expression of thenet force acting on the car.

  $F_{\text{net}}=F_s-F_d-F_r$   ........ (1)

Here, $F_s$ is the static friction force, $F_d$ is the drag force and $F_r$ is the rolling friction force applied on the car.

Write the expression for the rolling frictional force on the car.

  $$$F_r={\mu }_r\left(mg\right)$

Here, ${\mu }_r$ is the coefficient of rolling friction, $m$ is the mass of the car and $g$ is the acceleration due to gravity.

Substitute ${\mu }_r\left(mg\right)$ for $F_r$ , $C{v}^2$ for $F_d$ and $0$ for $F_{\text{net}}$ in expression (1).

  $F_s-C{v}^2-{\mu }_{r}mg=0$

Rearrange the above expression.

  $F_s=C{v}^2+{\mu }_{r}mg$   ........ (2)

Multiply both sides of the expression (2) by the constant speed $v$ of the car.

  $F_{s}v=\left(C{v}^2+{\mu }_{r}mg\right)v$

Substitute $P$ for $F_{s}v$ in the above expression.

  $P=\left(C{v}^2+{\mu }_{r}mg\right)v$   ........ (3)

Rearrange expression (3).

  $C=\frac{P}{v^3}-\frac{{\mu }_{r}mg}{v^2}$   ........ (4)

Calculation:

Substitute $13.5\text{hp}$ for $P$ , $55\text{mi}/\text{h}$ for $v$ , 0.0150 for ${\mu }_r$ , 1200 Kg for $m$ and $9.8{\text{ms}}^{\text{-2}}$ for $g$ in the expression(4).

  $\begin{array}{c}C=\frac{13.5\text{hp}}{{\left(55\text{mi}/\text{h}\right)}^3}-\frac{\left(0.0150\right)\left(1200\text{}\text{Kg}\right)\left(9.8{\text{ms}}^{\text{-2}}\right)}{{\left(55\text{mi}/\text{h}\right)}^2}\\ =\frac{13.5\text{hp}\left(\frac{746\text{W}}{1\text{hp}}\right)}{{\left(55\text{mi}/\text{h}\left(\frac{\left(1609.34\text{m}\right)\left(1h\right)}{\left(1\text{mi}\right)\left(3600\text{s}\right)}\right)\right)}^3}-\frac{\left(0.0150\right)\left(1200\text{}\text{Kg}\right)\left(9.8{\text{ms}}^{\text{-2}}\right)}{{\left(55\text{mi}/\text{h}\left(\frac{\left(1609.34\text{m}\right)\left(1h\right)}{\left(1\text{mi}\right)\left(3600\text{s}\right)}\right)\right)}^2}\\ =\frac{10071\text{W}}{14863.59{\text{m}}^{\text{3}}{\text{s}}^{\text{-3}}}-\frac{176.4\text{Kg}\text{m}{\text{s}}^{\text{-2}}}{604.53{\text{m}}^2{\text{s}}^{\text{-2}}}\\ =0.385\text{Kg}{\text{m}}^{\text{-1}}\end{array}$

Conclusion:

Thus, the value of $C$ is $0.385\text{Kg}{\text{m}}^{\text{-1}}$ .

(b)

To determine

The maximum speed of the car.

Explanation of Solution

Given:

The maximum power of the car is $164\text{hp}$ .

Calculation:

Write the expression of the net force acting on the car.

  $F_{\text{net}}=F_s-F_d-F_r$

Here, $F_s$ is the static friction force, $F_d$ is the drag force and $F_r$ is the rolling friction force applied on the car.

Write the expression for the rolling frictional force on the car.

  $$$F_r={\mu }_r\left(mg\right)$

Here, ${\mu }_r$ is the coefficient of rolling friction, $m$ is the mass of the car and $g$ is the acceleration due to gravity.

Substitute ${\mu }_r\left(mg\right)$ for $F_r$ , $C{v}^2$ for $F_d$ and $0$ for $F_{\text{net}}$ in expression (1).

  $F_s-C{v}^2-{\mu }_{r}mg=0$

Rearrange the above expression.

  $F_s=C{v}^2+{\mu }_{r}mg$

Multiply both sides of the expression (2) by the constant speed $v$ of the car.

  $F_{s}v=\left(C{v}^2+{\mu }_{r}mg\right)v$

Substitute $P$ for $F_{s}v$ in the above expression.

  $P=\left(C{v}^2+{\mu }_{r}mg\right)v$

Rearrange above expression.

  $v^3-\frac{{\mu }_{r}mg}{C}v-\frac{P}{C}=0$

Substitute $164\text{hp}$ for $P$ , 0.0150 for ${\mu }_r$ , 1200 Kg for $m$ , $0.385{\text{ms}}^{\text{-1}}$ for $P$ and $9.8{\text{ms}}^{\text{-2}}$ for $g$ in the above expression.

  $v^3-\frac{\left(0.0150\right)\left(1200\text{Kg}\right)\left(9.8{\text{ms}}^{\text{-2}}\right)}{0.385\text{Kg}{\text{m}}^{\text{-1}}}v-\frac{164\text{hp}}{0.385\text{Kg}{\text{m}}^{\text{-1}}}=0$

Simplify the above expression.

  $v^3-\left(\text{471}{\text{.4 m}}^{\text{2}}{\text{s}}^{\text{-2}}\right)v-\left(3.213\times {10}^5{\text{m}}^{\text{3}}{\text{s}}^{\text{-3}}\right)=0$

Using Graphine calculator to solve the above expression for $v$ .

  $\begin{array}{c}v=66.20{\text{ms}}^{\text{-1}}\\ =\left(66.20{\text{ms}}^{\text{-1}}\right)\left(\frac{0.6818\text{mi}{\text{h}}^{\text{-1}}}{0.3048{\text{ms}}^{\text{-1}}}\right)\\ =148\text{mi}{\text{h}}^{\text{-1}}\end{array}$

Conclusion:

Thus, the maximum speed of the car is $148\text{mi}/\text{h}$ .