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Chapter 3: Problem 55

A 900 -kg car is going \(20 \mathrm{~m} / \mathrm{s}\) along a level road. How large a constant retarding force is required to stop it in a distance of \(30 \mathrm{~m}\) ?

Short Answer

Expert verified
The required retarding force is \(-6000 \, \text{N}\).

Step by step solution

01

Identify Known Variables

We start by identifying the known variables in the problem. The mass of the car \( m \) is \( 900 \, \text{kg} \). Its initial speed \( v_i \) is \( 20 \, \text{m/s} \). The final velocity \( v_f \) is \( 0 \, \text{m/s} \) since the car is coming to a stop. The stopping distance \( d \) is \( 30 \, \text{m} \).
02

Use the Kinematic Equation to Find Acceleration

We'll use the kinematic equation: \[ v_f^2 = v_i^2 + 2ad \] to find the acceleration \( a \). Substituting the known values, we get: \[ 0 = 20^2 + 2a(30) \] which simplifies to \[ 400 = 60a \] leading to \[ a = \frac{400}{60} = \frac{20}{3} \, \text{m/s}^2 \] (negative because it is a retarding acceleration).
03

Calculate the Retarding Force

Use Newton's second law \( F = ma \) to find the force. Since the acceleration we found is retarding, the force is also negative, indicating its direction is opposite to the motion.\[ F = 900 \times \left( -\frac{20}{3} \right) \] Simplifying, we find \[ F = -6000 \, \text{N} \].

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Key Concepts

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

Understanding Kinematic Equations
Kinematic equations are fundamental in physics for describing the motion of objects. They relate key variables such as displacement, initial and final velocities, acceleration, and time.
In the problem, we have used the kinematic equation:
  • \[ v_f^2 = v_i^2 + 2ad \]
This equation helps us find the car's acceleration as it comes to a stop, given its initial velocity and the distance over which it stops.
  • \( v_f \) is the final velocity, which is \( 0 \, \text{m/s} \) for a stopping car.
  • \( v_i \) is the initial velocity, given as \( 20 \, \text{m/s} \).
  • \( a \) is the acceleration, which we need to find.
  • \( d \) is the displacement, the stopping distance of \( 30 \, \text{m} \).
By rearranging and solving the equation, we can find the necessary acceleration to stop the car within the given distance.
Newton's Second Law and Its Application
Newton's Second Law of Motion states that the force acting on an object is equal to the mass of that object times its acceleration. This can be expressed in the equation:
  • \[ F = ma \]
In the context of the problem, we use this law to calculate the necessary retarding force to halt the car. Here's how it works:
  • The mass \( m \) of the car is \( 900 \, \text{kg} \).
  • The acceleration \( a \) we found previously is negative \( -\frac{20}{3} \, \text{m/s}^2 \), indicating deceleration.
By multiplying these values, we find the retarding force. The negative sign of the force indicates it opposes the car's motion, a crucial insight for comprehending how forces interact in real-world scenarios.
This principle clarifies how forces act not only in halting cars but in many dynamics-related problems.
Role of Constant Retarding Force
A constant retarding force is one that remains the same throughout the motion and always opposes the direction of an object's movement. It is a crucial concept in bringing vehicles or objects to a stop efficiently.
  • The constant force means its magnitude does not change as the car slows down.
  • This type of force is essential as it ensures predictable and smooth braking over the intended distance.
In our solution, the calculated retarding force is \( -6000 \, \text{N} \). It is critical for ensuring that the car comes to a stop within the specified distance of \( 30 \, \text{m} \).
Understanding how a constant force acts helps in designing safety mechanisms, like brakes in a car, which need to apply consistent pressures to stop vehicles reliably. This concept not only applies to cars but to any object in motion where controlled deceleration is required.

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Most popular questions from this chapter

An elevator starts from rest with a constant upward acceleration. It moves \(2.0 \mathrm{~m}\) in the first \(0.60 \mathrm{~s}\). A passenger in the elevator is holding a \(3.0\) -kg package by a vertical string. What is the tension in the string during the accelerating process?

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

Consider an essentially spherical homogeneous celestial body of mass \(M\). The acceleration due to gravity in its vicinity beyond its surface at a distance \(R\) from its center is \(g_{R}\). Show that $$ g_{R}=\frac{G M}{R^{2}} $$ Notice that the acceleration drops off as \(1 / R^{2}\). Imagine an object of mass \(m\) at a distance \(R\) from the center of our celestial body. Its weight is \(F_{W}=m g_{R}\), but that's also the gravitation force on it due to the mass \(M\), that is, \(F_{W}=F_{G} .\) Hence, $$ m g_{R}=G \frac{m M}{R^{2}} $$

A \(5.0\) -kg mass hangs at the end of a cord. Find the tension in the cord if the acceleration of the mass is ( \(a\) ) \(1.5 \mathrm{~m} / \mathrm{s}^{2}\) up, (b) \(1.5 \mathrm{~m} / \mathrm{s}^{2}\) down, (c) \(9.81 \mathrm{~m} / \mathrm{s}^{2}\) down. Don't forget gravity.

The Moon, whose mass is \(7.35 \times 10^{22} \mathrm{~kg}\), orbits the Earth, whose mass is \(5.98 \times 10^{24} \mathrm{~kg}\), at a mean distance of \(3.85 \times 10^{8} \mathrm{~m}\). It is held in a nearly circular orbit by the Earth-Moon gravitational interaction. Determine the force of gravity due to the planet acting on the Moon. From the universal law of gravitation $$ F_{G}=G \frac{m M}{R^{2}} $$ we get $$ F_{G}=6.673 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg} \frac{\left(7.35 \times 10^{22} \mathrm{~kg}\right)\left(5.98 \times 10^{24}\right)}{\left(3.85 \times 10^{8} \mathrm{~m}\right)^{2}} $$ which yields $$ F_{G}=1.98 \times 10^{20} \mathrm{~N} $$ This is also the force on the Earth due to the Moon, and the force on the Moon due to the Earth.
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