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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 67.0 -kg person jumps from rest off a \(3.00-\mathrm{m}\) -high tower straight down into the water. Neglect air resistance. She comes to rest \(1.10 \mathrm{m}\) under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.

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 drawing shows two frictionless inclines that begin at ground level \((h=0 \mathrm{m})\) and slope upward at the same angle \(\theta\). One track is longer than the other, however. Identical blocks are projected up each track with the same initial speed \(v_{0} .\) On the longer track the block slides upward until it reaches a maximum height \(H\) above the ground. On the shorter track the block slides upward, flies off the end of the track at a height \(H_{1}\) above the ground, and then follows the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block is a height \(H_{2}\) above the end of the track. The initial total mechanical energy of each block is the same and is all kinetic energy. The initial speed of each block is \(v_{0}=7.00 \mathrm{m} / \mathrm{s},\) and each incline slopes upward at an angle of \(\theta=50.0^{\circ} .\) The block on the shorter track leaves the track at a height of \(H_{1}=1.25 \mathrm{m}\) above the ground. Find (a) the height \(H\) for the block on the longer track and (b) the total height \(H_{1}+H_{2}\) for the block on the shorter track.

A bicyclist rides 5.0 km due east, while the resistive force from the air has a magnitude of \(3.0 \mathrm{N}\) and points due west. The rider then turns around and rides \(5.0 \mathrm{km}\) due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of \(3.0 \mathrm{N}\) and points due east. (a) Find the work done by the resistive force during the round trip. (b) Based on your answer to part (a), is the resistive force a conservative force? Explain.

A 75.0-kg man is riding an escalator in a shopping mall. The escalator moves the man at a constant velocity from ground level to the floor above, a vertical height of \(4.60 \mathrm{m} .\) What is the work done on the man by (a) the gravitational force and (b) the escalator?
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