/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A \(1580-\mathrm{kg}\) car is tr... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(1580-\mathrm{kg}\) car is traveling with a speed of \(15.0 \mathrm{m} / \mathrm{s}\). What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of \(50.0 \mathrm{m} ?\)

Short Answer

Expert verified
The magnitude of the horizontal net force is 3555 N.

Step by step solution

01

Identify given values

The mass of the car (\( m \)) is \( 1580 \mathrm{~kg} \), the initial speed (\( v_i \)) is \( 15.0 \mathrm{~m/s} \), the final speed (\( v_f \)) is \( 0 \mathrm{~m/s} \) since the car is brought to a halt, and the stopping distance (\( d \)) is \( 50.0 \mathrm{~m} \).
02

Use kinematic equation to find acceleration

We use the kinematic equation \( v_f^2 = v_i^2 + 2a \, d \) to find acceleration (\( a \)). Substitute \( v_f = 0 \mathrm{~m/s} \), \( v_i = 15.0 \mathrm{~m/s} \), and \( d = 50.0 \mathrm{~m} \): \[ 0 = (15.0)^2 + 2a (50.0) \].
03

Solve for acceleration

Rearrange the equation to solve for \( a \): \[ 0 = 225 + 100a \]. Therefore, \( a = -\frac{225}{100} = -2.25 \mathrm{~m/s^2} \).
04

Use Newton's second law to find the force

Apply Newton's second law \( F = ma \) to find the force. Substitute the mass \( m = 1580 \mathrm{~kg} \) and acceleration \( a = -2.25 \mathrm{~m/s}^2 \): \[ F = 1580 \times (-2.25) = -3555 \mathrm{~N} \].
05

Determine the magnitude of the force

The magnitude of the force is the absolute value of the calculated force: \( |F| = 3555 \mathrm{~N} \).

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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 a set of four equations that can predict unknown values like displacement, velocity, time, and acceleration. These equations apply only to objects moving with constant acceleration. In our exercise, bringing a car to a halt utilizes the kinematic equation:
  • \( v_f^2 = v_i^2 + 2a d \)
This equation lets us find acceleration when a car comes to a stop over a certain distance. Here, we use it by knowing:
  • Initial speed \( (v_i) = 15.0 \mathrm{~m/s} \)
  • Final speed \( (v_f) = 0 \mathrm{~m/s} \)
  • Distance \((d) = 50.0 \mathrm{~m}\)
We plug these into the equation:\[0 = (15.0)^2 + 2a \, (50.0)\]Solving gives us the required acceleration. The negative sign in acceleration indicates deceleration, meaning the car slows down.
Newton's Second Law
Newton's second law of motion is a fundamental principle connecting force, mass, and acceleration. It is expressed as:
  • \( F = ma \)
Where:
  • \( F \) is the net force applied to an object
  • \( m \) is the mass of the object
  • \( a \) is the acceleration
In our problem, we use this law to find the force needed to stop the car. The given mass is \( 1580 \mathrm{~kg} \), and the acceleration calculated is \( -2.25 \mathrm{~m/s}^2 \).Plugging these values into the formula:\[F = 1580 \times (-2.25) = -3555 \, \mathrm{N}\]This result tells us the force magnitude, and the negative sign signifies direction, emphasizing that the force is opposite to the car's motion.
Net Force Calculation
Net force is the total force acting on an object, considering all individual forces. It's crucial when determining how an object accelerates or decelerates. In the exercise, after finding the acceleration, we calculated the net force using what Newton described as the relationship between mass and acceleration. We determined the force required to stop the car over a set distance.The magnitude of this force is critical for understanding practical applications like brake design, ensuring a vehicle can safely and effectively stop. The calculated net force here is \( |3555 \, \mathrm{N}| \), helping us understand the stopping power needed. This concept illustrates how various forces interact and impacts our everyday life, especially in transportation and safety engineering.

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Most popular questions from this chapter

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of \(151 \mathrm{N}\) on the wire. The left section of the wire makes an angle of \(14.0^{\circ}\) relative to the horizontal and sustains a tension of \(447 \mathrm{N} .\) Find the magnitude and direction of the tension that the right section of the wire sustains.

A woman stands on a scale in a moving elevator. Her mass is 60.0 kg, and the combined mass of the elevator and scale is an additional 815 kg. Starting from rest, the elevator accelerates upward. During the acceleration, the hoisting cable applies a force of \(9410 \mathrm{N}\). What does the scale read during the acceleration?

Part \(a\) of the drawing shows a bucket of water suspended from the pulley of a well; the tension in the rope is \(92.0 \mathrm{N}\). Part \(b\) shows the same bucket of water being pulled up from the well at a constant velocity. What is the tension in the rope in part \(b ?\)

On earth, two parts of a space probe weigh \(11000 \mathrm{N}\) and \(3400 \mathrm{N}\). These parts are separated by a center-to-center distance of \(12 \mathrm{m}\) and may be treated as uniform spherical objects. Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other objects.

A bowling ball (mass \(=7.2 \mathrm{kg}\), radius \(=0.11 \mathrm{m}\) ) and a billiard ball (mass \(=0.38 \mathrm{kg}\), radius \(=0.028 \mathrm{m}\) ) may each be treated as uniform spheres. What is the magnitude of the maximum gravitational force that each can exert on the other?

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