91Ó°ÊÓ

A \(600-\mathrm{kg}\) 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 \(v_{i x}=30 \mathrm{~m} / \mathrm{s}, v_{f x}=0\), and \(x=70 \mathrm{~m}\). Use \(v_{f x}^{2}=v_{i x}^{2}+2 a x\) to find $$ a=\frac{v_{f x}^{2}-v_{i x}^{2}}{2 x}=\frac{0-900 \mathrm{~m}^{2} / \mathrm{s}^{2}}{140 \mathrm{~m}}=-6.43 \mathrm{~m} / \mathrm{s}^{2} $$ 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 \(\mu_{s}=F_{\mathrm{f}} / F_{N}\), 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, $$ F_{N}=F_{W}=m g=(600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5886 \mathrm{~N} $$ so that $$ \mu_{s}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{3858}{5886}=0.66 $$ The coefficient of friction must be at least \(0.66\) if the car is to stop within \(70 \mathrm{~m}\).

Step by step solution

01

Identify Given Values

We are provided with the following values: the mass of the car, \( m = 600 \, \text{kg} \); the initial velocity of the car, \( v_{i} = 30 \, \text{m/s} \); the final velocity of the car, \( v_{f} = 0 \, \text{m/s} \); and the stopping distance, \( x = 70 \, \text{m} \).

02

Use the Kinematic Equation to Find Acceleration

We use the equation \( v_{f}^2 = v_{i}^2 + 2ax \) to find the acceleration \( a \). Substituting the known values: \( 0 = (30)^2 + 2a(70) \). Solving for \( a \), we get \( a = \frac{0 - 900}{140} = -6.43 \, \text{m/s}^2 \).

03

Calculate the Retarding Force

Using Newton's second law, \( F = ma \), we find the retarding force. Substitute \( m = 600 \, \text{kg} \) and \( a = -6.43 \, \text{m/s}^2 \): \( F = 600 \times (-6.43) = -3858 \, \text{N} \). The negative sign indicates the force is opposite the direction of motion.

04

Determine the Normal Force on the Car

The normal force \( F_N \) is equal to the gravitational force on the car since the road is level. So, \( F_N = mg = 600 \times 9.81 = 5886 \, \text{N} \).

05

Calculate the Minimum Coefficient of Friction

The minimum coefficient of static friction \( \mu_s \) is given by \( \mu_s = \frac{F_f}{F_N} \), where \( F_f = 3858 \, \text{N} \) is the frictional force. Thus, \( \mu_s = \frac{3858}{5886} = 0.66 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

kinematic equations

Kinematic equations are essential in physics to describe the motion of objects. They help us understand how velocity, acceleration, and displacement interrelate over time. These equations are particularly useful in solving problems involving uniformly accelerated motion, like the ones in this exercise.

In part (a) of the exercise, we used the equation: \[ v_{f}^2 = v_{i}^2 + 2ax \]This equation allows us to find the acceleration when the initial and final velocities, as well as the displacement, are known. Here, the initial velocity \( v_i \) was 30 m/s, the final velocity \( v_f \) was 0 m/s, and the stopping distance \( x \) was 70 m. Substituting these values helps to find acceleration \( a \) necessary to stop the car in the given distance.

Understanding which kinematic equation to apply depends on what aspects of motion you are analyzing. In situations where time is not a factor, this particular equation becomes very handy.

Newton's second law

Newton's second law provides a fundamental connection between force and motion. It states that the force acting on an object is equal to the mass of the object times its acceleration, given by the formula:\[ F = ma \]This principle was used to find the retarding force needed to stop the car.

Once we calculated the negative acceleration using the kinematic equation, we applied Newton's second law to determine the force required. The car mass \( m \) given is 600 kg, and the acceleration \( a \) was found to be \(-6.43 \text{ m/s}^2\). This allowed us to calculate the retarding force \( F \) as \(-3858 \text{ N} \), indicating that the force acts contrary to the direction of motion to bring the car to a stop.

Using Newton's second law is crucial for understanding how forces govern motion. It helps in calculating unknowns like force when mass and acceleration are known, providing insights into dynamics of moving objects.

static friction

Static friction is the frictional force that prevents surfaces from sliding past each other. It acts when an object is at rest relative to a surface. This type of friction is crucial when calculating the forces required to stop a vehicle without the tires slipping on the road.

In part (b) of the exercise, static friction comes into play since the problem specifies that the wheels are not locked and there is no sliding. The static frictional force must be equal to the necessary retarding force, which was calculated to be \( 3858 \text{ N} \). This means \( F_f = 3858 \text{ N} \), and this force is what prevents the car from moving until the external force surpasses it.

Static friction is usually characterized by a coefficient, expressed as \( \mu_s \), which is the ratio of the maximum static frictional force to the normal force. This coefficient is vital in ensuring stability and safety in vehicles' stopping capabilities under various conditions.

retarding force

Retarding force, often synonymous with stopping force, is the force applied to slow down or stop a moving object. It acts in the opposite direction of the object's motion, effectively reducing its speed over time.

In the context of the problem, the retarding force was needed to bring the car to a stop over a distance of 70 meters. By using the kinematic equation and Newton's second law, this force was found to be \(-3858 \text{ N} \). The negative sign here signifies the force works against the initial direction of the car's movement.

Retarding forces are critical in braking mechanisms and other scenarios where controlled deceleration is necessary. Understanding these forces helps in designing systems that safely reduce speed and ensure vehicles stop within desired distances.

coefficient of friction

The coefficient of friction is a numerical value that represents the degree of interaction between two surfaces in contact. It varies based on the materials and surfaces involved, affecting how much frictional force can develop.

In this problem, we calculated the minimum static coefficient of friction \( \mu_s \) needed to stop the car within the given distance. The frictional force \( F_f \) was found to be \( 3858 \text{ N} \) and the normal force \( F_N \) was \( 5886 \text{ N} \), giving a coefficient:\[ \mu_s = \frac{F_f}{F_N} = 0.66 \]This value means that the tire-road interface must provide sufficient grip to accomplish the stop without slipping.

The coefficient of friction is crucial for safety in vehicle dynamics. It determines how effectively a car can decelerate under normal road conditions, essential for ensuring driver control and stability during braking.