/*! 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 7 A person pushes horizontally wit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 7

A person pushes horizontally with a force of \(220 \mathrm{~N}\) on a \(55 \mathrm{~kg}\) crate to move it across a level floor. The coefficient of kinetic friction between the crate and the floor is \(0.35 .\) What is the magnitude of (a) the frictional force and (b) the acceleration of the crate?

Short Answer

Expert verified
The frictional force is approximately 188.84 N, and the acceleration is 0.57 m/s虏.

Step by step solution

01

Calculate the Normal Force

The normal force \( F_n \) is equal to the weight of the crate, which is given by the formula: \( F_n = m \cdot g \), where \( m = 55\, \mathrm{kg} \) is the mass of the crate and \( g = 9.81\, \mathrm{m/s^2} \) is the acceleration due to gravity. Therefore, \( F_n = 55 \cdot 9.81 = 539.55\, \mathrm{N} \).
02

Calculate the Frictional Force

The frictional force \( F_f \) can be calculated using the formula: \( F_f = \mu_k \cdot F_n \), where \( \mu_k = 0.35 \) is the coefficient of kinetic friction. Substitute the values: \( F_f = 0.35 \cdot 539.55 = 188.8425\, \mathrm{N} \). Thus, the magnitude of the frictional force is approximately \( 188.84\, \mathrm{N} \).
03

Calculate the Net Force

The net force \( F_{net} \) exerted on the crate is the applied force minus the frictional force. Given the horizontal push \( F_{push} = 220\, \mathrm{N} \), the net force is: \( F_{net} = F_{push} - F_f = 220 - 188.84 = 31.16\, \mathrm{N} \).
04

Calculate the Acceleration

Use Newton's second law to find the acceleration \( a \) of the crate: \( F_{net} = m \cdot a \). Thus, \( a = \frac{F_{net}}{m} = \frac{31.16}{55} = 0.566\, \mathrm{m/s^2} \). The crate's acceleration is approximately \( 0.57\, \mathrm{m/s^2} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 helps us understand how forces affect the motion of an object. According to the law, the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. This relationship is captured by the famous equation \( F = m \cdot a \), where \( F \) represents the net force, \( m \) is the mass, and \( a \) is the acceleration. In simple terms:
  • If you push harder (increase the force), the object accelerates more.
  • If the object is heavier (more mass), it accelerates less for the same force.
In our exercise, once the net force on the crate was determined, we applied Newton's second law to calculate its acceleration. The net force took into account both the applied force and the opposing frictional force.
Normal force calculation
The normal force is an essential concept when discussing motion on surfaces. It is the force exerted perpendicular or "normal" to the contact surface. For an object resting on a flat surface, like our crate, the normal force supports the object's weight. It is calculated using the object's mass and gravitational force.
Formula for normal force on a horizontal surface is given by: \( F_n = m \cdot g \), where:
  • \( m \) is the mass of the object.
  • \( g \) is the acceleration due to gravity \( (9.81 \, \mathrm{m/s^2}) \).
In our problem, the mass was \( 55 \, \mathrm{kg} \), leading to a normal force of \( 539.55 \, \mathrm{N} \). Normal force is crucial because the frictional force depends on it. Without calculating it, we couldn't determine how much friction was opposing the motion.
Net force determination
Calculating the net force is a critical step in understanding an object's motion. Net force is essentially the sum of all forces acting on an object, considering their direction. If forces are balanced, the net force is zero, resulting in no change in motion. In our exercise, the net force is the result of a push force and a frictional force working against this push.
The steps to determine the net force include:
  • Identifying all forces acting on the object: In our scenario, a horizontal push and frictional force.
  • Finding magnitudes and directions of these forces.
  • Subtracting the forces that oppose the motion from those that support it: here, \( F_{net} = 220 \, \mathrm{N} - 188.84 \, \mathrm{N} = 31.16 \, \mathrm{N} \).
This net force is what ultimately affects the acceleration, allowing us to apply Newton's second law and find the precise acceleration value for the crate.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 cat dozes on a stationary merry-go-round in an amusement park, at a radius of \(5.4 \mathrm{~m}\) from the center of the ride. Then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every \(6.0 \mathrm{~s}\). What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding (or the cat clinging with its claws)?

In Fig. \(6-39,\) a car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill, the normal force on the driver from the car seat is \(0 .\) The driver's mass is \(70.0 \mathrm{~kg}\). What is the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley?

Calculate the ratio of the drag force on a jet flying at \(1000 \mathrm{~km} / \mathrm{h}\) at an altitude of \(10 \mathrm{~km}\) to the drag force on a propdriven transport flying at half that speed and altitude. The density of air is \(0.38 \mathrm{~kg} / \mathrm{m}^{3}\) at \(10 \mathrm{~km}\) and \(0.67 \mathrm{~kg} / \mathrm{m}^{3}\) at \(5.0 \mathrm{~km} .\) Assume that the airplanes have the same effective cross-sectional area and drag coefficient \(C\).
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Circular Motion and Gravitation

Read Explanation

Famous Physicists

Read Explanation

Modern Physics

Read Explanation

Particle Model of Matter

Read Explanation

Work Energy and Power

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.