/*! 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 20 A 16-kg sled is being pulled alo... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 20

A 16-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of \(24 \mathrm{N}\). Starting from rest, the sled attains a speed of \(2.0 \mathrm{m} / \mathrm{s}\) in \(8.0 \mathrm{m} .\) Find the coefficient of kinetic friction between the runners of the sled and the snow.

Short Answer

Expert verified
The coefficient of kinetic friction is approximately 0.128.

Step by step solution

01

Calculate Acceleration

First, we need to calculate the acceleration of the sled. Using the equation of motion: \[ v^2 = u^2 + 2as \]Where: \( v = 2.0 \) m/s (final speed), \( u = 0 \) m/s (initial speed), and \( s = 8.0 \) m (distance). \[ (2.0)^2 = 0 + 2a \times 8.0 \]\[ 4.0 = 16a \]\[ a = 0.25 \text{ m/s}^2 \]
02

Apply Newton's Second Law

Using Newton's second law, sum the forces acting on the sled: \[ F_{ ext{net}} = ma \]Where \( F_{ ext{net}} \) is the net horizontal force, \( m = 16 \) kg, and \( a = 0.25 \) m/s虏. The pulling force is 24 N and friction acts opposite to this force. \[ F_{ ext{net}} = F_{ ext{pull}} - F_{ ext{friction}} \]\[ 16 \times 0.25 = 24 - F_{ ext{friction}} \]\[ 4.0 = 24 - F_{ ext{friction}} \]\[ F_{ ext{friction}} = 20 \text{ N} \]
03

Find Normal Force and Coefficient of Friction

Since the surface is horizontal, the normal force \( F_N \) equals the gravitational force acting on the sled, which is \( mg \). \[ F_N = mg = 16 \times 9.8 = 156.8 \text{ N} \]The coefficient of kinetic friction \( \mu_k \) is given by:\[ F_{ ext{friction}} = \mu_k F_N \]Substitute \( F_{ ext{friction}} = 20 \) N:\[ 20 = \mu_k \times 156.8 \]\[ \mu_k = \frac{20}{156.8} \approx 0.128 \]

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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 is a fundamental principle in physics. It tells us how the motion of an object changes when forces are applied. In simple terms, it says that the force acting on an object is equal to the mass of the object multiplied by its acceleration. This can be written as the equation: \[ F = ma \]Where:- \( F \) is the force,- \( m \) is the mass of the object,- \( a \) is the acceleration.In our sled problem, Newton's Second Law helps us understand how the sled's speed changes. The sled is pulled by a force, which makes it accelerate. The net force on the sled is the pulling force minus the friction force. Using Newton's Second Law, we calculate this net force as mass times acceleration. This step is crucial in figuring out the friction force that opposes the motion.
Equation of Motion
The equation of motion is a key tool in physics to predict the behavior of moving objects. It involves formulas that relate speed, time, distance, and acceleration. They are particularly handy when calculating how an object moves over time.In this scenario, we use the equation:\[ v^2 = u^2 + 2as \]Where:- \( v \) is the final velocity,- \( u \) is the initial velocity,- \( a \) is the acceleration,- \( s \) is the distance traveled.This equation helps us find the sled's acceleration from rest to a final speed over a certain distance. By substituting the known values, we solve for the unknown acceleration. This value is crucial for other calculations like determining the net force using Newton's second law.
Normal Force
The normal force is the support force exerted by a surface, perpendicular to the object resting on it. When an object, such as the sled, is on a horizontal surface, the normal force is equal to the gravitational force pulling it down. This means it acts in the opposite direction to gravity and balances it out.In our sled problem, the normal force \( F_N \) can be calculated using the equation:\[ F_N = mg \]Where:- \( m \) is the mass of the sled,- \( g \) is the acceleration due to gravity (approximately \(9.8 \text{ m/s}^2\)).Once we know the normal force, we can use it to find the coefficient of kinetic friction, which is the ratio of the friction force to the normal force. This coefficient tells us how slippery or rough the surface is and is calculated using the friction force obtained from Newton鈥檚 second law.

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Most popular questions from this chapter

A pendulum consists of a small object hanging from the ceiling at the end of a string of negligible mass. The string has a length of \(0.75 \mathrm{m}\). With the string hanging vertically, the object is given an initial velocity of \(2.0 \mathrm{m} / \mathrm{s}\) parallel to the ground and swings upward on a circular arc. Eventually, the object comes to a momentary halt at a point where the string makes an angle \(\theta\) with its initial vertical orientation and then swings back downward. Find the angle \(\theta\)

A \(35-\mathrm{kg}\) girl is bouncing on a trampoline. During a certain interval after she leaves the surface of the trampoline, her kinetic energy decreases to 210 J from 440 J. How high does she rise during this interval? Neglect air resistance.

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of \(83.0 \mathrm{kg}\), and the height of the water slide is \(11.8 \mathrm{m} .\) If the kinetic frictional force does \(-6.50 \times 10^{3} \mathrm{J}\) of work, how fast is the student going at the bottom of the slide?

The (nonconservative) force propelling a \(1.50 \times 10^{3}-\mathrm{kg}\) car up a mountain road does \(4.70 \times 10^{6} \mathrm{J}\) of work on the car. The car starts from rest at sea level and has a speed of \(27.0 \mathrm{m} / \mathrm{s}\) at an altitude of \(2.00 \times 10^{2} \mathrm{m}\) above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are nonconservative forces.

A 75.0-kg skier rides a 2830-m-long lift to the top of a mountain. The lift makes an angle of \(14.6^{\circ}\) with the horizontal. What is the change in the skier's gravitational potential energy?
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