/*! 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 109 You push a box along the floor a... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 109

You push a box along the floor against a constant force of friction. When you push with a horizontal force of \(75 \mathrm{N}\), the acceleration of the box is \(0.50 \mathrm{m} / \mathrm{s}^{2} ;\) when you increase the force to \(81 \mathrm{N},\) the acceleration is \(0.75 \mathrm{m} / \mathrm{s}^{2} .\) Find (a) the mass of the box and (b) the coefficient of kinetic friction between the box and the floor.

Short Answer

Expert verified
The box's mass is 24 kg, and the kinetic friction coefficient is approximately 0.27.

Step by step solution

01

Understand the Problem

You are given the forces applied to a box and the resulting accelerations. Two scenarios are provided. We need to find the mass of the box and the coefficient of kinetic friction.
02

Apply Newton's Second Law for the First Scenario

For the first scenario, apply Newton's second law: \[ F_{net} = ma \]where \( F_{net} \) is the net force on the box, \( m \) is the mass, and \( a \) is the acceleration. The net force is the difference between the applied force and the frictional force: \[ F_{net} = 75 \, N - f_k \]With \( a = 0.50 \, m/s^2 \), substitute the known values:\[ 75 \, N - f_k = m \times 0.50 \, m/s^2 \]
03

Apply Newton's Second Law for the Second Scenario

For the second scenario, again use:\[ F_{net} = ma \]The net force becomes:\[ F_{net} = 81 \, N - f_k \]With the new acceleration \( a = 0.75 \, m/s^2 \), substitute the values:\[ 81 \, N - f_k = m \times 0.75 \, m/s^2 \]
04

Solve the System of Equations

We now have two equations:1. \( 75 \, N - f_k = m \, \times 0.50 \, m/s^2 \)2. \( 81 \, N - f_k = m \, \times 0.75 \, m/s^2 \)Subtract the first equation from the second:\[ (81 - 75) \, N = m \times (0.75 - 0.50) \, m/s^2 \]\[ 6 = 0.25m \]Solve for \( m \):\[ m = \frac{6}{0.25} = 24 \text{ kg} \]
05

Calculate the Kinetic Friction Force

Using \( m = 24 \text{ kg} \) in one of the original equations to find \( f_k \). Use equation 1:\[ 75 \, N - f_k = 24 \, kg \times 0.50 \, m/s^2 \]\[ 75 \, N - f_k = 12 \text{ N} \]\[ f_k = 75 \, N - 12 \, N = 63 \, N \]
06

Calculate the Coefficient of Kinetic Friction

The force of kinetic friction \( f_k \) is given by:\[ f_k = \mu_k \times m \times g \]where \( \mu_k \) is the coefficient of kinetic friction and \( g \) is the acceleration due to gravity \( 9.8 \, m/s^2 \).\[ 63 \, N = \mu_k \times 24 \, kg \times 9.8 \, m/s^2 \]Solve for \( \mu_k \):\[ \mu_k = \frac{63}{24 \times 9.8} \approx 0.27 \]
07

Conclude the Results

The mass of the box is \( 24 \, kg \) and the coefficient of kinetic friction \( \mu_k \) is approximately \( 0.27 \).

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Key Concepts

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

Force of Friction
When you push an object along a surface, you're always met with a resisting force called friction. Friction helps objects like a box or a car slow down when there's a push or pull involved. This resistance is due to the contact between the surfaces鈥攍ike the box against the floor in this scenario. Frictional force acts in the opposite direction of the motion. This means, when you are pushing a box forward, the force of friction is pulling it in the opposite direction.

In physics, frictional forces are divided into two broad categories: static friction and kinetic friction. However, in this case, we're dealing with kinetic friction since the box is moving. Friction that happens during motion is known as kinetic friction, and its magnitude can be calculated using specific formulas. The frictional force is proportional to the force pressing the two surfaces together (normal force). In more technical terms, the frictional force (\( f_k \)) is given by:
  • \( f_k = \, \mu_k \times N \)
  • where\( N \)is the normal force which, for horizontal motion, equals the weight of the object rendered through gravity.
This relationship is crucial to finding other variables, such as the coefficient of kinetic friction.
Coefficient of Kinetic Friction
The coefficient of kinetic friction, denoted by \( \mu_k \), is a dimensionless number that represents the ratio between the kinetic frictional force and the normal force acting on an object. Simply put, it describes how much friction is there when two objects slide over each other. Different surface materials affect this coefficient. For instance, rubber on concrete will have a much higher \( \mu_k \) than ice on ice.

In the given problem, the coefficient of kinetic friction is part of the equation: \[ f_k = \mu_k \times m \times g \]where:
  • \( f_k \)is the kinetic frictional force you already calculated to be \(63 \, N\).
  • \( m \)is mass, determined to be \(24 \, kg\)
  • \( g \)is the acceleration due to gravity, approximately \(9.8 \, m/s^2\).
By solving this equation, you find that \( \mu_k \approx 0.27 \). This number helps in understanding the frictional characteristics between our box and the floor.
System of Equations
In the context of solving physics problems like this, a system of equations is a set of two or more equations with the same variables. The system is collectively used to solve for unknowns鈥攊n our case, the mass of the box and the frictional force. Let's break down why this is important.

In this particular scenario, we have been given different forces and their corresponding accelerations. Thus, each condition forms its own equation based on Newton's Second Law, \( F_{net} = ma \). From the provided information:
  • For \(75 \, N\)force: \( 75 \, N - f_k = m \times 0.50 \, m/s^2 \)
  • For \(81 \, N\)force: \( 81 \, N - f_k = m \times 0.75 \, m/s^2 \)
Solving this system involves manipulating these equations to isolate and find the unknowns. In our equations, by subtracting one from another, we eliminate \( f_k \)and end up solving for mass first. Then, backtrack using one equation to determine the force of friction. This approach, known as simultaneous equations, is a standard method in physics to solve for multiple unknowns in interlinked problems.

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Most popular questions from this chapter

Force Times Distance I At the local hockey rink, a puck with a mass of \(0.12 \mathrm{kg}\) is given an initial speed of \(v=5.3 \mathrm{m} / \mathrm{s}\) (a) If the coefficient of kinetic friction between the ice and the puck is 0.11 , what distance \(d\) does the puck slide before coming to rest? (b) If the mass of the puck is doubled, does the frictional force \(F\) exerted on the puck increase, decrease, or stay the same? Explain. (c) Does the stopping distance of the puck increase, decrease, or stay the same when its mass is doubled? Explain. (d) For the situation considered in part (a), show that \(F d=\frac{1}{2} m v^{2}\). The significance of this result will be discussed in Chapter 7 , where we will see that \(\frac{1}{2} m v^{2}\) is the kinetic energy of an object.)

Find the coefficient of kinetic friction between a \(3.85-\mathrm{kg}\) block and the horizontal surface on which it rests if an \(850-\mathrm{N} / \mathrm{m}\) spring must be stretched by \(6.20 \mathrm{cm}\) to pull it with constant speed. Assume that the spring pulls in the horizontal direction.

An object moves on a flat surface with an acceleration of constant magnitude. If the acceleration is always perpendicular to the object's direction of motion, (a) is the shape of the object's path circular, linear, or parabolic? (b) During its motion, does the object's velocity change in direction but not magnitude, change in magnitude but not direction, or change in both magnitude and direction? (c) Does its speed increase, decrease, or stay the same?

Two drivers traveling side-by-side at the same speed suddenly see a deer in the road ahead of them and begin braking. Driver 1 stops by locking up his brakes and screeching to a halt; driver 2 stops by applying her brakes just to the verge of locking, so that the wheels continue to turn until her car comes to a complete stop. (a) All other factors being equal, is the stopping distance of driver 1 greater than, less than, or equal to the stopping distance of driver \(2 ?\) (b) Choose the best explanation from among the following: I. Locking up the brakes gives the greatest possible braking force. II. The same tires on the same road result in the same force of friction. III. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.

A 97 -kg sprinter wishes to accelerate from rest to a speed of \(13 \mathrm{m} / \mathrm{s}\) in a distance of \(22 \mathrm{m}\). (a) What coefficient of static friction is required between the sprinter's shoes and the track? (b) Explain the strategy used to find the answer to part (a).
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