91Ó°ÊÓ

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 \(70 \mathrm{~m}\).

Step by step solution

01

Determine Initial and Final Velocities

We begin by identifying the initial velocity \( v_{i} \) and the final velocity \( v_{f} \). The car starts at \( v_{i} = 30 \) m/s and stops at \( v_{f} = 0 \) m/s.

02

Use Equation for Constant Acceleration

We use the equation \( v_{f}^2 = v_{i}^2 + 2ax \) to find the acceleration \( a \). Substituting the known values: \( (0)^2 = (30)^2 + 2a(70) \).

03

Solve for Acceleration

Rearranging the equation, we get: \[ a = \frac{0 - 900}{2 \times 70} = -6.43 \, \text{m/s}^2 \]. This is the deceleration required to stop the car.

04

Calculate Retarding Force Using Newton's Second Law

The retarding force \( F \) is calculated using \( F = ma \). Here \( m = 600 \) kg and \( a = -6.43 \, \text{m/s}^2 \), giving \[ F = 600 \times -6.43 = -3858 \, \text{N} \].

05

Identify the Normal Force

The normal force \( F_N \) acting on the car is equal to its weight, \( F_N = mg \), where \( g = 9.81 \, \text{m/s}^2 \). Thus, \( F_N = 600 \times 9.81 = 5886 \, \text{N} \).

06

Calculate Minimum Coefficient of Static Friction

The minimum coefficient of static friction \( \mu_s \) is calculated using \( \mu_s = \frac{F_f}{F_N} \), where \( F_f = 3858 \, \text{N} \). Therefore, \[ \mu_s = \frac{3858}{5886} = 0.66 \].

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Key Concepts

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

Retarding Force

In the world of physics, understanding the concept of a retarding force is crucial when analyzing a stopping or slowing object. The retarding force is the force applied opposite to the direction of motion, leading an object to slow down or stop. In the context of the 600-kg car, this force had to be determined to stop the car within 70 meters.

To calculate this force, we can use Newton’s Second Law of Motion, which states that the force acting on an object is equal to the mass of the object times its acceleration. This is expressed as \( F = ma \). In our example, since the car is slowing down, the acceleration is actually a deceleration and is negative, indicating a reduction in speed. Calculating, we identified that the retarding force was \(-3858 \text{ N} \). This negative sign denotes that the force is working against the direction of travel, which is why it is termed as 'retarding.'

To bring this to a practical understanding, think about gently applying brakes on a bicycle. The resistance you feel is essentially the retarding force slowing you down.

Coefficient of Friction

When friction enters the picture, it often relates to the surface interaction between two objects, and that's where the coefficient of friction comes in. This coefficient is a number that represents the frictional forces between two bodies. It’s crucial when determining if one object can move against another.

In our scenario, the coefficient of friction was needed to determine how effectively the car's tires could grip the road and stop it in 70 meters. This coefficient can be computed when the retarding force is already known by using the formula \( \mu = \frac{F_f}{F_N} \), where \( F_f \) is the frictional force and \( F_N \) is the normal force. In this exercise, it turned out to be 0.66, meaning a relatively moderate amount of friction is needed to stop the car without skidding.

Understanding this concept can also help anticipate conditions where the tires may not effectively stop the car, such as on icy or wet surfaces where the coefficient is much lower.

Constant Acceleration

Constant acceleration is a fundamental concept of motion. It refers to a steady increase or decrease in velocity, meaning every second, the speed changes by the same amount. It is a vital principle when analyzing problems involving linear motion since it simplifies calculations with predictable results.

In the given exercise, the car experienced a constant deceleration (negative acceleration) as it came to a halt. The formula used was \( v_f^2 = v_i^2 + 2ax \), known as the equation for constant acceleration. Calculations showed that this deceleration was \(-6.43 \text{ m/s}^2 \). This means that every second, the car's speed decreases by 6.43 meters per second until it stops entirely.

For constant acceleration concepts, always ensure you are using the correct direction for acceleration (positive for speeding up, negative for slowing down), as this influences your final results.

Static Friction

Static friction is the force that keeps an object at rest when it is subject to an external force. It acts between surfaces when there is no relative movement between them. This is crucial when we assume the wheels are not locked, implying no sliding occurs.

In the car problem, static friction is what enables the car to have a grip on the road instead of sliding when brakes are applied. This type of friction is involved before the tires begin to move relative to the road's surface. The maximum static friction can be calculated using \( F_f = \mu_s \cdot F_N \). If this force exceeds, the tires would begin to slide, moving into the kinetic friction realm.

Importantly, static friction is always equal to or greater than the force being applied up until the point of motion, ensuring the object remains stationary or, like the car, stops without skidding. This makes understanding and calculating static friction crucial for vehicle safety and understanding how forces interact on stationary objects.