/*! 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 79 A truck is traveling at \(11.1 \... [FREE SOLUTION] | 91Ó°ÊÓ

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A truck is traveling at \(11.1 \mathrm{~m} / \mathrm{s}\) down a hill when the brakes on all four wheels lock. The hill makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is \(0.750\). How far does the truck skid before coming to a stop?

Short Answer

Expert verified
The truck skids approximately 19.45 meters before stopping.

Step by step solution

01

Identify Forces Involved

First, identify all the forces acting on the truck. The gravitational force acting along the incline is \( mg \sin \theta \), the normal force is \( mg \cos \theta \), and the force of kinetic friction is \( f_k = \mu_k N = \mu_k mg \cos \theta \).
02

Calculate Net Force

The net force acting on the truck along the incline is the difference between the component of gravity along the incline and the friction force. So, \( F_{\text{net}} = mg \sin \theta - f_k = mg \sin \theta - \mu_k mg \cos \theta \).
03

Use Newton's Second Law

Apply Newton's second law to find acceleration. Since \( F = ma \), we have \( mg \sin \theta - \mu_k mg \cos \theta = ma \). Simplifying, \( a = g (\sin \theta - \mu_k \cos \theta) \).
04

Substitute Given Values

Substitute the known values into the equation for acceleration: \( g = 9.8 \mathrm{~m/s^2} \), \( \theta = 15.0^{\circ} \), and \( \mu_k = 0.750 \). Calculate \( a = 9.8 \times (\sin 15.0^{\circ} - 0.750 \cos 15.0^{\circ}) \).
05

Calculate Acceleration

First, compute \( \sin 15.0^{\circ} \approx 0.2588 \) and \( \cos 15.0^{\circ} \approx 0.9659 \). Thus, \( a = 9.8 \times (0.2588 - 0.750 \times 0.9659) \). Calculate \( a \approx -3.168 \mathrm{~m/s^2} \) indicating deceleration.
06

Use Kinematic Equation to Find Distance

Use the kinematic equation \( v^2 = u^2 + 2as \) where final velocity \( v = 0 \), initial velocity \( u = 11.1 \mathrm{~m/s} \), and solve for distance \( s \): \( 0 = (11.1)^2 + 2(-3.168)s \).
07

Solve for Skid Distance

Rearrange the equation to \( s = \frac{(11.1)^2}{2 \times 3.168} \). Compute \( s \approx 19.45 \mathrm{~m} \).

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Key Concepts

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

Newton's Second Law
Newton's Second Law of Motion is fundamental in understanding how objects behave under the influence of forces. Simply put, it states that the force acting on an object is equal to the mass of that object times its acceleration, expressed as the formula:
  • \( F = ma \)
In the context of our problem, this law helps us calculate the acceleration of the truck as it slides down the incline. Different forces influence the truck, including gravitational force and friction. By applying Newton's Second Law, we can evaluate these forces to find the net force that dictates the acceleration of the truck.
Determining this acceleration is crucial because it shows how fast the truck slows down, leading us to calculate the distance it travels before coming to a stop. This practical application of Newton's Second Law is essential for solving problems involving motion and forces.
Kinematic Equations
Kinematic equations describe the motion of objects and allow us to find variables such as distance, time, velocity, and acceleration without needing to consider forces directly. One of the most commonly used kinematic equations in problems like ours is:
  • \( v^2 = u^2 + 2as \)
In this equation,
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity,
  • \( a \) is the acceleration, and
  • \( s \) is the distance traveled.

In our scenario, we know that the truck comes to a stop, so the final velocity \( v = 0 \). The initial velocity \( u \) is provided as 11.1 m/s, and the acceleration \( a \) has been determined using Newton's Second Law. Plugging these values into our kinematic equation allows us to solve for \( s \), the total distance the truck skids down the hill. Understanding how to use these equations is a key skill in physics, enabling us to describe motion quantitatively.
Gravitational Force
Gravitational force is one of the fundamental forces acting on objects near Earth's surface. It is the force that gives us weight and affects any object on an incline. In our inclined plane problem, gravitational force can be broken down into two components:
  • A component acting parallel to the plane, \( mg \sin \theta \), which causes the truck to slide down, and
  • A component acting perpendicular to the plane, \( mg \cos \theta \), which is balanced by the normal force.
Understanding these components is crucial as they influence the net force experienced by the truck. The gravitational force component along the incline adds to the truck's motion, while the perpendicular component helps calculate the normal force, which in turn assists in finding the frictional force acting against the truck's movement. Exploring gravitational force helps deepen our understanding of how forces operate in real-world scenarios, particularly on inclines.
Inclined Plane
An inclined plane, or a slope, is one of the simplest types of machines and is frequently analyzed in physics for educational purposes. It offers a practical setting to study force components and motion. When an object is placed on an inclined plane, like our truck, forces acting upon it must be resolved into parallel and perpendicular components relative to the plane.
These include the gravitational force components and the force of kinetic friction. The angle of the incline, \( \theta \), plays a significant role as it affects both the gravitational component in the plane's direction and the opposing friction force. It also determines the net acceleration that influences the truck's skidding distance.
Understanding inclined planes and calculating how forces distribute in such scenarios allows for better insights into motion and provides practical knowledge often applied in engineering and technology fields.

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Most popular questions from this chapter

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