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Chapter 6: Problem 7

A person pushes horizontally with a force of 260 N on a 55 kg crate to move it across a level floor. The coefficient of kinetic friction is 0.30. What is the magnitude of (a) the frictional force and (b) the crate鈥檚 acceleration?

Short Answer

Expert verified
(a) The frictional force is 161.7 N. (b) The crate's acceleration is approximately 1.79 m/s虏.

Step by step solution

01

Calculate the Normal Force

Since the crate is on a horizontal surface and there is no vertical acceleration, the normal force is equal to the weight of the crate. The weight of the crate can be calculated using the formula:\[ F_{\text{weight}} = m \cdot g \]where \( m = 55 \text{ kg} \) and \( g = 9.8 \, \text{m/s}^2 \). Thus:\[ F_{\text{weight}} = 55 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 539 \, \text{N} \]So, the normal force \( N \) is 539 N.
02

Calculate the Frictional Force

The frictional force can be calculated using the formula:\[ f_{\text{friction}} = \mu_k \cdot N \]where \( \mu_k = 0.30 \) is the coefficient of kinetic friction and \( N = 539 \, \text{N} \) is the normal force. Therefore:\[ f_{\text{friction}} = 0.30 \times 539 \, \text{N} = 161.7 \, \text{N} \]
03

Calculate the Net Force

The net force acting on the crate is the difference between the applied force and the frictional force:\[ F_{\text{net}} = F_{\text{applied}} - f_{\text{friction}} \]Given \( F_{\text{applied}} = 260 \, \text{N} \) and \( f_{\text{friction}} = 161.7 \, \text{N} \), we find:\[ F_{\text{net}} = 260 \, \text{N} - 161.7 \, \text{N} = 98.3 \, \text{N} \]
04

Calculate the Crate鈥檚 Acceleration

Using Newton's second law, the acceleration can be found as follows:\[ F_{\text{net}} = m \cdot a \]where \( m = 55 \, \text{kg} \). Solving for \( a \) gives:\[ a = \frac{F_{\text{net}}}{m} = \frac{98.3 \, \text{N}}{55 \, \text{kg}} \approx 1.788 \, \text{m/s}^2 \]

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Key Concepts

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

Kinetic Friction
When moving objects slide against each other, kinetic friction comes into play, working to resist their motion. Kinetic friction depends on two primary factors: the nature of the surfaces in contact and the normal force pressing them together. Here, the coefficient of kinetic friction, \(\mu_k\), is a measure of how rough or smooth two surfaces are. In this problem, it is given as 0.30, indicating that the floor and the crate have moderate friction. The force of kinetic friction can be found using the formula:
  • \( f_{\text{friction}} = \mu_k \cdot N \)
where \( N \) is the normal force. This force acts opposite to the direction of motion, making the push required to move the crate larger.
Normal Force
The normal force is crucial to understanding how forces interact on surfaces. It's the force exerted by a surface to support the weight of an object resting on it, acting perpendicular to the surface. In our scenario with the crate resting on a flat floor, the normal force balances the gravitational force pulling the crate downward, thus:
  • \( N = F_{\text{weight}} = m \cdot g \)
Given that the crate has a mass of \( 55 \text{ kg} \) and gravity \( g \) is \( 9.8 \text{ m/s}^2 \), we find that the normal force \( N \) is 539 N. The normal force is pivotal in calculating kinetic friction, as it directly influences the frictional force magnitude.
Net Force
The net force is the sum of all forces acting on an object, accounting for both magnitude and direction. It's the overall force that causes an acceleration according to Newton's Second Law.To find the net force on the crate, subtract the frictional force from the applied force:
  • \( F_{\text{net}} = F_{\text{applied}} - f_{\text{friction}} \)
In this problem, you've applied 260 N to move the crate, but friction (161.7 N) acts against this motion, leaving a net force of 98.3 N. This is the effective force propelling the crate.
Acceleration
Acceleration tells us how quickly an object's speed changes. It is directly related to the net force and inversely related to the object's mass, as articulated by Newton's Second Law. This relationship is spelled out in the equation:
  • \( F_{\text{net}} = m \cdot a \)
For the acceleration \( a \) of the crate, we rearrange the formula to find:
  • \( a = \frac{F_{\text{net}}}{m} \)
Plugging in the values, \( F_{\text{net}} = 98.3 \text{ N} \) and \( m = 55 \text{ kg} \), gives an acceleration of approximately 1.788 \text{ m/s}^2. This means the crate speeds up by \( 1.788 \text{ m/s} \) each second when you push it across the floor.

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Most popular questions from this chapter

A roller-coaster car has a mass of 1300 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 20 m, its speed is not changing. At the top of the hill, what are the (a) magnitude FN and (b) direction (up or down) of the normal force on the car from the track if the car鈥檚 speed is v 11 m/s? What are (c) FN and (d) the direction if v 14 m/s?

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