Chapter 6, Problem 46SP

A 1200-kg car coasts from rest down a driveway that is inclined 20° to the horizontal and is 15 m long. How fast is the car going at the end of the driveway if (a) friction is negligible and (b) a friction force of 3000 N opposes the motion?

To determine

The speed of the car, of $1200\text{ kg}$ mass, at the end of the driveway that is inclined $20°$ to the horizontal and is $\text{15 m}$ long, considering the frictional force is negligible.

Answer to Problem 46SP

Solution:

$10\text{ m/s}$

Explanation of Solution

Given data:

The mass of the car is $1200\text{ kg}$.

The angle of incline of the driveway with the horizontal is $20°$.

The length of the driveway is $15\text{ m}$.

Formula used:

The expression for change in kinetic energy is written as

$KE=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Here, $m$ is the mass, $v_f$ is the final velocity, $v_i$ is the initial velocity, and $KE$ is the change in kinetic energy.

The expression for change in potential energy is written as

$P{E}_G=mg\left(h_f-h_i\right)$

Here, $g$ is the acceleration due to gravity, $h_f$ is the final height, $h_i$ is the initial height, and $P{E}_G$ is the change in potential energy.

The expression for work energy theorem is written as

$\text{Change in kinetic energy}+\text{change in potential energy}=\text{Work done}$

Explanation:

Draw the diagram for the motion of the car

There is no frictional force acting on the car, so there is no work done on the car. Therefore,

$\begin{array}{c}\text{Change in kinetic energy}+\text{change in potential energy}=\text{0}\\ KE+P{E}_G=0\end{array}$

Substitute $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $KE$ and $mg\left(h_f-h_i\right)$ for $P{E}_G$

$\begin{array}{c}\frac{1}{2}m\left(v_f^2-v_i^2\right)+mg\left(h_f-h_i\right)=0\\ \frac{1}{2}{v}_f^2+g{h}_f=\frac{1}{2}{v}_i^2+g{h}_i\end{array}$

The car started from rest. The initial velocity is zero. Also, the final position of the car is on the horizontal. Therefore,

$h_f=0$

Consider the diagram and the initial vertical distance covered is the sine component of the distance covered. Therefore,

$\begin{array}{c}{h}_i=\left(15\text{ m}\right)\sin 20°\\ =5.13\text{ m}\end{array}$

Substitute $5.13\text{ m}$ for $h_i$, $0\text{ m}$ for $h_f$, $0\text{ m/s}$ for $v_i$, and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}\frac{1}{2}{v}_f^2+\left(9.81{\text{ m/s}}^2\right)\left(0\text{ m}\right)=\frac{1}{2}{\left(0\text{ m/s}\right)}^2+\left(9.81{\text{ m/s}}^2\right)\left(5.13\text{ m}\right)\\ v_f=\sqrt{\left(2\right)\left(50.3253\right)}\\ v_f=10.03\text{ m/s}\\ {\text{v}}_f\simeq 10\text{ m/s}\end{array}$

Conclusion:

The speed of the car at the end of the driveway with negligible frictional force is $10\text{ m/s}$.

To determine

The speed of the car, of $1200\text{ kg}$ mass, at the end of the driveway that is inclined $20°$ to the horizontal and is $\text{15 m}$ long, considering the frictional force of $3000\text{ N}$.

Answer to Problem 46SP

Solution:

$5.1\text{ m/s}$

Explanation of Solution

Given data:

The mass of the car is $1200\text{ kg}$.

The angle of incline is $20°$.

The distance to the horizontal is $15\text{ m}$.

The frictional force is $3000\text{ N}$.

Formula used:

The expression of change in kinetic energy is written as

$KE=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

Here, $m$ is the mass, $v_f$ is the final velocity, $v_i$ is the initial velocity, and $KE$ is the change in kinetic energy.

The expression of change in potential energy is written as

$P{E}_G=mg\left(h_f-h_i\right)$

Here, $g$ is the acceleration due to gravity, $h_f$ is the final height, $h_i$ is the initial height, and $P{E}_G$ is the change in potential energy.

The expression of frictional work done is written as

$W=F_{f}s\cos \theta$

Here, $F_f$ is the frictional force, $W$ is the work done, $s$ is the displacement, and $\theta$ is the angle between the displacement and force.

The expression for the work energy theorem is written as

$\text{Change in kinetic energy}+\text{change in potential energy}=\text{Work done}$

Explanation:

Draw the diagram for the motion of the car

Recall that the expression for the work energy theorem is written as

$\begin{array}{c}\text{Change in kinetic energy}+\text{change in potential energy}=\text{Work done}\\ KE+P{E}_G=W\end{array}$

Substitute $\frac{1}{2}m\left(v_f^2-v_i^2\right)$ for $KE$, $F_{f}s\cos \theta$ for $W$, and $mg\left(h_f-h_i\right)$ for $P{E}_G$

$\frac{1}{2}m\left(v_f^2-v_i^2\right)+mg\left(h_f-h_i\right)=F_{f}s\cos \theta$

The car started from rest. The initial velocity is zero. Also, the final position of the car is on the horizontal. Therefore,

$h_f=0$

Consider the diagram and the initial vertical distance covered is the sine component of the distance covered. Therefore,

$\begin{array}{c}{h}_i=\left(15\text{ m}\right)\sin 20°\\ =5.13\text{ m}\end{array}$

The frictional force acts in the opposite direction of the displacement. Therefore, $\theta =180°$.

Substitute $5.13\text{ m}$ for $h_i$, $0\text{ m}$ for $h_f$, $0\text{ m/s}$ for $v_i$, $180°$ for $\theta$, $3000\text{ N}$ for $F_f$, $1200\text{ kg}$ for $m$, $15\text{ m}$ for $s$, and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}\frac{1}{2}\left(1200\text{ kg}\right)\left(v_f^2-{\left(0\text{ m/s}\right)}^2\right)+\left(1200\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)\left(0\text{ m}-5.13\text{ m}\right)=\left(3000\text{ N}\right)\left(15\text{ m}\right)\cos 180°\\ 600v_f^2-60390.36=-45000\\ v_f=\sqrt{\frac{60390.36-45000}{600}}\\ v_f\simeq 5.1\text{ m/s}\end{array}$

Conclusion:

The speed of the car with a frictional force of $3000\text{ N}$ is $5.1\text{ m/s}$.