A 600 -kg car is coasting along a level road at 30 m / s. (a) How large a retarding force (assu

A 600 -kg car is coasting along a level road at 30 m / s....

A 600 -kg car is coasting along a level road at $30 \mathrm{~m} / \mathrm{s}$. (a) How large a retarding force (assumed constant) is required to stop it in a distance of $70 \mathrm{~m} ?(b)$ What is the minimum coefficient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction- there's no sliding. ( $a$ ) First find the car's acceleration from a constant- $a$ equation. It is known that $\mathrm{v}_{i x}=30 \mathrm{v}_{f x},=0$, and $x=70 \mathrm{~m}$. Use $v_{k}^{2}=v_{u}^{2}+2 a x$ to find $$ a=\frac{v_{f x}^{2}-v_{i r}^{2}}{2 x}=\frac{0-900 \mathrm{~m}^{2} / \mathrm{s}^{2}}{140 \mathrm{~m}}=-6.43 \mathrm{~m} / \mathrm{s}^{2} $$ Now write $$ F=m a=(600 \mathrm{~kg})\left(-6.43 \mathrm{~m} / \mathrm{s}^{2}\right)-3858 \mathrm{~N} \text { or }-3.9 \mathrm{kN} $$ (b) Assume the force found in $(a)$ is supplied as the friction force between the tires and roadway. Therefore, the magnitude of the friction force on the tires is $F_{\mathrm{f}}=3858 \mathrm{~N}$. The coefficient of friction is given by $u=F_{f} / F_{N}$, where $F_{N}$ is the normal force. In the present case, the roadway pushes up on the car with a force equal to the car's weight. Therefore, that $$ \begin{array}{l} F_{N}=F_{W}=m g=(600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5886 \mathrm{~N} \\ \mu_{s}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{3858}{5886}=0.66 \end{array} $$ The coefficient of friction must be at least $0.66$ if the car is to stop within $70 \mathrm{~m}$.

A 600 -kg car is coasting along a level road at $30 \mathrm{~m} / \mathrm{s}$. (a) How large a retarding force (assumed constant) is required to stop it in a distance of $70 \mathrm{~m} ?(b)$ What is the minimum coefficient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction- there's no sliding.
( $a$ ) First find the car's acceleration from a constant- $a$ equation. It is known that $\mathrm{v}_{i x}=30 \mathrm{v}_{f x},=0$, and $x=70 \mathrm{~m}$. Use $v_{k}^{2}=v_{u}^{2}+2 a x$ to
find
$$
a=\frac{v_{f x}^{2}-v_{i r}^{2}}{2 x}=\frac{0-900 \mathrm{~m}^{2} / \mathrm{s}^{2}}{140 \mathrm{~m}}=-6.43 \mathrm{~m} / \mathrm{s}^{2}
$$
Now write
$$
F=m a=(600 \mathrm{~kg})\left(-6.43 \mathrm{~m} / \mathrm{s}^{2}\right)-3858 \mathrm{~N} \text { or }-3.9 \mathrm{kN}
$$
(b) Assume the force found in $(a)$ is supplied as the friction force between the tires and roadway. Therefore, the magnitude of the friction force on the tires is $F_{\mathrm{f}}=3858 \mathrm{~N}$. The coefficient of friction is given by $u=F_{f} / F_{N}$, where $F_{N}$ is the normal force. In the present case, the roadway pushes up on the car with a force equal to the car's weight. Therefore,
that
$$
\begin{array}{l}
F_{N}=F_{W}=m g=(600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5886 \mathrm{~N} \\
\mu_{s}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{3858}{5886}=0.66
\end{array}
$$
The coefficient of friction must be at least $0.66$ if the car is to stop within $70 \mathrm{~m}$.

Key Concepts

Normal Force

The normal force is the perpendicular contact force exerted by a surface on an object in contact with it. On a level surface, the normal force is typically equal to the weight of the object and is a critical factor when calculating frictional forces using the coefficient of friction.

Friction Force

Friction force is the resistive force that opposes the relative motion or tendency of such motion between two surfaces in contact. In the context of braking, friction provides the force necessary to decelerate a vehicle by acting between the tires and the road surface.

Coefficient of Friction

The coefficient of friction is a dimensionless scalar that quantifies the ratio of the frictional force to the normal force between two surfaces. It indicates how much frictional force is available to resist motion, and determines whether braking can be accomplished without sliding.

Deceleration

Deceleration is a specific type of acceleration where the velocity decreases over time. It is calculated as a negative acceleration and is essential when analyzing the requirements for safely stopping a moving object within a certain distance.

Uniformly Accelerated Motion

This concept involves motion with constant acceleration. Equations of uniformly accelerated motion relate displacement, initial velocity, final velocity, acceleration, and time. In problems involving deceleration, these equations allow us to compute the acceleration needed to change the velocity over a known distance.

Newton's Second Law of Motion

Newton's Second Law states that the net force acting on an object is equal to the product of its mass and acceleration (F = m * a). This principle is used to determine the magnitude of the force required to achieve a given acceleration or deceleration, given the object's mass.

Transcript

0:00 Hi.

00:01 In this problem we have a car whose mass is equal to 600 kilograms and this car is coasting or moving with an initial velocity of 30 meters per second.

00:18 The question here in part a, what is the minimum retarding force that is required to stop this car in a distance of 70 meters? so here we have a distance of 70 meters, and we require the car to stop completely in that distance, so the given velocity is equal to zero, in other words.

00:43 Now, let's use this equation of kinematics, which is vf squared equal to the initial velocity squared, plus 2a multiplied by the distance.

00:56 So from here, we can substitute with the final velocity being equal to zero, and in that way we will have that the final velocity squared is equal to negative because of switching sides.

01:16 And here we have two multiplied by the acceleration, multiplied by the distance, which is 70 meters.

01:24 From this equation we can rearrange it to find out that the acceleration is equal to 30 squared divided by minus 2 times 70 which is equal to negative 6 .4 3 so that's the acceleration or the deceleration to be accurate because of the negative sign that is required to stop this car in that given distance.

01:55 Now let's use the equation of newton's second law that relates the force to the mass of the object and its acceleration so that we can find the retarding force that is required to produce this acceleration.

02:10 So from here, substituting with the mass with 600 multiplied by negative 6 .43 and that's equal to 3 ,800.

02:29 Or approximately 3 .9 kiloons...

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