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Chapter 6: Problem 2

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 \(3.5 \mathrm{~kg}\) book is pushed from rest through a distance of \(0.90 \mathrm{~m}\) by the horizontal \(25 \mathrm{~N}\) force from the broom and then has a speed of \(1.60 \mathrm{~m} / \mathrm{s}\), what is the coefficient of kinetic friction between the book and floor?

Short Answer

Expert verified
The coefficient of kinetic friction is approximately 0.58.

Step by step solution

01

Identify Given Variables

We have the following known values from the problem:- Mass of the calculus book, \( m = 3.5 \text{ kg} \).- Force applied by the broom, \( F = 25 \text{ N} \).- Distance over which the force is applied, \( d = 0.90 \text{ m} \).- Final speed of the book, \( v = 1.60 \text{ m/s} \).- Initial speed of the book, \( u = 0 \text{ m/s} \), since it starts from rest.
02

Calculate Work Done by Applied Force

The work done by the force is given by the formula:\[ W = F \times d = 25 \text{ N} \times 0.90 \text{ m} = 22.5 \text{ J}. \]
03

Calculate Kinetic Energy Acquired

The kinetic energy \( KE \) acquired by the book is given by:\[ KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 3.5 \text{ kg} \times (1.60 \text{ m/s})^2 = 4.48 \text{ J}. \]
04

Calculate Work Done Against Friction

The work done against friction \( W_f \) is the difference between the work done by the broom and the kinetic energy gained:\[ W_f = W - KE = 22.5 \text{ J} - 4.48 \text{ J} = 18.02 \text{ J}. \]
05

Calculate Force of Friction

The work done against friction is also given by \( W_f = f_k \times d \), where \( f_k \) is the force of kinetic friction:\[ f_k = \frac{W_f}{d} = \frac{18.02 \text{ J}}{0.90 \text{ m}} = 20.02 \text{ N}. \]
06

Determine Coefficient of Kinetic Friction

The force of kinetic friction is also given by \( f_k = \mu_k \times m \times g \), where \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Solving for \( \mu_k \) gives:\[ \mu_k = \frac{f_k}{m \times g} = \frac{20.02 \text{ N}}{3.5 \text{ kg} \times 9.8 \text{ m/s}^2} \approx 0.58. \]

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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, stating that the force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, it's expressed as \( F = m imes a \). This concept helps us understand how the broom acts on the calculus book in the exercise. Here:
  • The force \( F \) is given as 25 N (newtons).
  • The mass \( m \) of the book is 3.5 kg.
  • The initial acceleration can be deduced using these values.
To find acceleration \( a \), rearrange the formula to \( a = \frac{F}{m} \). By plugging in the given values, \( a = \frac{25 \text{ N}}{3.5 \text{ kg}} \approx 7.14 \text{ m/s}^2 \).
This acceleration tells us how quickly the book is speeding up (owing to the broom's force), directly involving Newton's Second Law in deciphering how this force translates into motion.
Work-Energy Principle
The Work-Energy Principle is a core concept that connects the ideas of work and kinetic energy. In simple terms, it states that the work done on an object is equal to the change in its kinetic energy. In our shuffleboard example, the broom does work on the book by pushing it.
  • The work done by the broom's force can be calculated using the formula: \( W = F \times d \), where \( F \) is the force and \( d \) is the distance over which the force is applied.
  • Here, \( W = 25 \text{ N} \times 0.90 \text{ m} = 22.5 \text{ J} \).
  • This work results in a change in the kinetic energy of the book.
Thus, using this principle, we can link the mechanical work to the resultant movement of the book, providing a better grasp of how energy is transferred and transformed during the game.
Kinetic Energy Calculation
Kinetic energy is the energy that an object possesses due to its motion. The formula to calculate kinetic energy is \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity of the object.
  • In our example, the mass \( m \) of the book is 3.5 kg, and the final velocity \( v \) is 1.60 m/s.
  • Substituting these values into the formula, we get \( KE = \frac{1}{2} \times 3.5 \text{ kg} \times (1.60 \text{ m/s})^2 \approx 4.48 \text{ J} \).
This calculation shows how the book's kinetic energy increases due to the force exerted by the broom. The gain in kinetic energy is a practical representation of its speed once the force is applied. It also helps verify the work-energy principle by comparing with the work done earlier.
Friction Force Calculation
Friction is the force resisting the motion of the book across the hallway surface. The frictional force plays a crucial role in calculating the coefficient of kinetic friction, \( \mu_k \). To find the friction force \( f_k \), we consider the difference in work done by the broom and the kinetic energy acquired:
  • The work done against friction \( W_f \) is the difference: \( W_f = W - KE = 22.5 \text{ J} - 4.48 \text{ J} = 18.02 \text{ J} \).
  • The friction force is calculated as \( f_k = \frac{W_f}{d} = \frac{18.02 \text{ J}}{0.90 \text{ m}} = 20.02 \text{ N} \).
  • Now, using \( f_k = \mu_k \times m \times g \) where \( g = 9.8 \text{ m/s}^2 \), solving for \( \mu_k \) gives \( \mu_k = \frac{f_k}{m \times g} = \frac{20.02 \text{ N}}{3.5 \text{ kg} \times 9.8 \text{ m/s}^2} \approx 0.58 \).
Understanding the frictional forces helps in understanding how they affect the book's movement, especially in slowing it down after the applied force.

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Most popular questions from this chapter

A block of mass \(m_{t}=4.0 \mathrm{~kg}\) is put on top of a block of mass \(m_{b}=5.0 \mathrm{~kg}\). To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12 N must be applied to the top block. The assembly of blocks is now placed on a horizontal, frictionless table (Fig. \(6-47\) ). Find the magnitudes of (a) the maximum horizontal force \(\vec{F}\) that can be applied to the lower block so that the blocks will move together and (b) the resulting acceleration of the blocks.

A baseball player with mass \(m=79 \mathrm{~kg}\), sliding into second base, is retarded by a frictional force of magnitude \(470 \mathrm{~N}\). What is the coefficient of kinetic friction \(\mu_{k}\) between the player and the ground?

A bedroom bureau with a mass of \(45 \mathrm{~kg}\), including drawers and clothing, rests on the floor. (a) If the coefficient of static friction between the bureau and the floor is \(0.45\), what is the magnitude of the minimum horizontal force that a person must apply to start the bureau moving? (b) If the drawers and clothing, with \(17 \mathrm{~kg}\) mass, are removed before the bureau is pushed, what is the new minimum magnitude?

Two blocks, of weights \(3.6 \mathrm{~N}\) and \(7.2 \mathrm{~N}\), are connected by a massless string and slide down a \(30^{\circ}\) inclined plane. The coefficient of kinetic friction between the lighter block and the plane is \(0.10\), and the coefficient between the heavier block and the plane is 0.20. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the taut string.

A block slides with constant velocity down an inclined plane that has slope angle \(\theta\). The block is then projected up the same plane with an initial speed \(v_{0}\). (a) How far up the plane will it move before coming to rest? (b) After the block comes to rest, will it slide down the plane again? Give an argument to back your answer.
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