/*! 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 2 In a pickup game of dorm shuffle... [FREE SOLUTION] | 91Ó°ÊÓ

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In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 3.5 kg book is pushed from rest through a distance of 1.20 m by the horizontal 25 N force from the broom and then has a speed of 1.75 m/s, what is the coefficient of kinetic friction between the book and floor?

Short Answer

Expert verified
The coefficient of kinetic friction is approximately 0.599.

Step by step solution

01

Understand the Problem

We have a 3.5 kg book pushed by a 25 N force over 1.20 m. It reaches a speed of 1.75 m/s. We need to find the coefficient of kinetic friction between the book and the floor.
02

Calculate Work Done by the Broom

The work done by the broom, which is the force times the distance, is given by the formula: \( W = F \times d \). Substituting the values, we get: \( W = 25 \text{ N} \times 1.20 \text{ m} = 30 \text{ J} \).
03

Calculate Kinetic Energy of the Book

The kinetic energy of the book when it reaches the speed of 1.75 m/s is given by \( KE = \frac{1}{2}mv^2 \). Substituting for mass \( m = 3.5 \text{ kg} \) and velocity \( v = 1.75 \text{ m/s} \), we get: \( KE = \frac{1}{2} \times 3.5 \times (1.75)^2 \approx 5.359375 \text{ J} \)
04

Calculate Work Done Against Friction

The work done against friction is the difference between the work done by the broom and the kinetic energy: \( W_{friction} = W - KE \). Therefore, \( W_{friction} = 30 \text{ J} - 5.359375 \text{ J} = 24.640625 \text{ J} \).
05

Determine the Friction Force

Since work done against friction \( W_{friction} \) is equal to the friction force multiplied by the distance, \( f_k \times d = W_{friction} \). Solving for friction force \( f_k \), we have: \( f_k = \frac{24.640625}{1.20} \approx 20.534 \text{ N} \).
06

Calculate the Coefficient of Kinetic Friction

Using the friction force \( f_k = \mu_k \times N \), where \( N \) is the normal force. Here, \( N = mg = 3.5 \text{ kg} \times 9.8 \text{ m/s}^2 = 34.3 \text{ N} \). Therefore, the coefficient of friction \( \mu_k \) is \( \mu_k = \frac{f_k}{N} = \frac{20.534}{34.3} \approx 0.599 \).

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a foundational concept in physics. It states that the work done on an object is equal to the change in its kinetic energy. This principle helps us understand the relationship between the forces acting on an object and the object's motion.
In our shuffleboard exercise, the work-energy principle is used to calculate the energy changes as the book slides along the hallway. The work done by the broom is calculated as the force applied by the broom multiplied by the distance the book travels. This gives us an initial total work of 30 joules.
Understanding this principle allows us to relate the initial work done to the final kinetic energy of the book, which helps in determining the energy lost to friction as the book moves, allowing us to solve for the coefficient of kinetic friction.
Coefficient of Friction
The coefficient of friction is a value that represents the frictional force between two surfaces. It varies depending on the materials of the surfaces in contact. In this problem, we are concerned with the kinetic coefficient of friction, which applies when objects are moving relative to each other.
This coefficient is determined by dividing the frictional force by the normal force. The value is dimensionless and typically ranges from 0 (perfectly smooth, no friction) to greater than 1 (very rough surface).
Here, after calculating the work done against friction and the subsequent force of friction, we use the normal force to find the kinetic coefficient of friction. It's essential for understanding how much force is needed to overcome the friction between the sliding book and the floor.
Normal Force
Normal force is a contact force that acts perpendicular to the surface an object is resting upon. It counteracts the weight of the object, preventing it from accelerating downwards due to gravity.
In our example, the normal force is equal to the weight of the book since it is resting on a horizontal surface. The weight is determined by multiplying the mass of the book by the acceleration due to gravity, which we approximate as 9.8 m/s². Thus, the normal force is 34.3 N for the book.
This force is crucial when calculating the coefficient of friction, as it directly influences the frictional force the book encounters as it slides across the hallway floor.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It is calculated using the formula: \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity.
In the shuffleboard activity, the book gains kinetic energy as it is pushed by the broom. When reaching a velocity of 1.75 m/s, it acquires a kinetic energy of about 5.36 joules.
This measure of kinetic energy is essential for understanding how much energy is being transferred from the work done by the broom into motion, as well as how much energy is lost due to frictional forces, helping us solve the puzzle of finding the coefficient of kinetic friction.
Newton's Laws
Newton's Laws of Motion govern all motion, including that of the calculus book in our experiment. Particularly, the first law (law of inertia) and the second law (F=ma) are relevant here.
The first law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by a net external force. In our scenario, this external force is the broom pushing the book, overcoming the initial state of rest.
The second law provides the foundation for calculating the acceleration due to the broom's force, even though friction opposes it. By connecting force, mass, and acceleration, we can understand how quickly the book speeds up initially and how friction then slows it down, confirming the interrelation between force, motion, and energy.

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Most popular questions from this chapter

A puck of mass m 1.50 kg slides in a circle of radius r 25.0 cm on a frictionless table while attached to a hanging cylinder of mass M 2.50 kg by means of a cord that extends through a hole in the table (Fig. 6-45). What speed keeps the cylinder at rest?

Suppose the coefficient of static friction between the road and the tires on a car is 0.60 and the car has no negative lift.What speed will put the car on the verge of sliding as it rounds a level curve of 32.0 m radius?

A water droplet 4.0 mm in diameter is falling with a speed of 10 km/h at an altitude of 20 km. Another droplet 6.0 mm in diameter is falling at 25% of that speed and at 25% of that altitude. The density of air at 20 km is 0.20 kg/m3 and that at 5.0 km is 0.70 kg/m3 . Assume that the drag coefficient C is the same for the two drops. Find the ratio of the drag force on the higher drop to that on the lower drop.

The mysterious sliding stones. Along the remote Racetrack Playa in Death Valley, California, stones sometimes gouge out prominent trails in the desert floor, as if the stones had been migrating (Fig. 6-16). For years curiosity mounted about why the stones moved. One explanation was that strong winds during occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about 0.80. What horizontal force must act on a 20 kg stone (a typical mass) to maintain the stone’s motion once a gust has started it moving? (Story continues with Problem 37.)

What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 35 km/h and the ms between tires and track is 0.40?

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