/*! 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 6 A person pushes a 16.0-kg shoppi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 6

A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of \(22.0 \mathrm{m}\). She pushes in a direction \(29.0^{\circ}\) below the horizontal. A \(48.0-\mathrm{N}\) frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.

Short Answer

Expert verified
The shopper exerts a force of approximately 54.7 N, does 1046.6 J of work, friction does -1056 J of work, and gravity does 0 J of work.

Step by step solution

01

Understanding the Scenario

The shopping cart is moving at constant velocity, which means the net force acting on it is zero. The shopper is pushing down at an angle below the horizontal, and a frictional force is opposing the motion.
02

Analyzing Forces

Since the cart is moving at a constant velocity, the horizontal component of the pushing force must equal the frictional force. Let \( F \) be the magnitude of the force the shopper exerts.The horizontal component of \( F \) is \( F \cos(29^{\circ}) \). We know:\[ F \cos(29^{\circ}) = 48.0 \text{ N} \]
03

Calculating the Pushing Force

Solve for \( F \) using the equation from Step 2:\[ F = \frac{48.0}{\cos(29^{\circ})} \]Calculating gives:\[ F \approx 54.7 \text{ N} \]
04

Calculating the Work Done by the Pushing Force

Work is calculated by the formula:\[ \text{Work} = F \cdot d \cdot \cos(\theta) \]Substitute the known values:\[ \text{Work} = 54.7 \text{ N} \times 22.0 \text{ m} \times \cos(29^{\circ}) \]This simplifies to approximately:\[ \text{Work} \approx 1046.6 \text{ J} \]
05

Calculating the Work Done by the Frictional Force

The work done by friction is given by:\[ \text{Work} = -f \cdot d \]Substitute the known values:\[ \text{Work} = -48.0 \text{ N} \times 22.0 \text{ m} \]\[ \text{Work} = -1056.0 \text{ J} \]
06

Work Done by the Gravitational Force

Since the gravitational force acts vertically and there is no vertical motion of the cart, the work done by gravity is:\[ \text{Work} = 0 \text{ J} \]

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.

Constant Velocity
The concept of constant velocity is crucial in understanding the motion of the shopping cart. When an object moves with a constant velocity, it means that its speed and direction do not change over time. This implies that the forces acting on the object are perfectly balanced, resulting in no acceleration.
In the context of the shopping cart, moving at constant velocity indicates that the total net force is zero.
  • The horizontal component of the force pushing the cart is balanced by the frictional force.
  • There is no acceleration in any direction, maintaining a uniform motion.
Understanding constant velocity simplifies the analysis of forces, as it implies that the shopper's exerted force horizontally matches the resistance provided by friction.
Force Components
To fully grasp how forces contribute to the cart's movement, we need to discuss force components. When a force is applied at an angle, its effects need to be analyzed in two orthogonal directions: horizontal and vertical. This is done using trigonometry.
In this scenario:
  • The force applied by the shopper is dissected into horizontal and vertical components.
  • The horizontal component, which aligns with the cart's direction, can be calculated as \( F \cos(29^{\circ}) \).
  • This component is crucial as it directly counters the frictional force of 48.0 N.
  • The vertical component is \( F \sin(29^{\circ}) \), which affects the load on the cart but does not contribute to horizontal motion.
Analyzing force components helps us identify how much force is available for the actual movement and how much affects other aspects like load and balance.
Frictional Force
Frictional force plays a key role in the motion of the cart, as it opposes the direction of movement. It is caused by the contact between the wheels and the surface, resisting the cart's sliding. In this specific case:
  • The frictional force given is 48.0 N, acting against the horizontal movement.
  • This force helps us determine the necessary horizontal component of the applied force.
  • By balancing against this frictional force, we can maintain constant velocity, as there's no net horizontal force.
Frictional force is not only a resistance; understanding it also allows calculations on required force exertion for any given movements at a given constant speed.
Gravitational Force
Gravitational force is the force exerted by Earth's gravity on any object with mass. While it is constant in our daily lives, its effects vary based on the object's motion and orientation.
  • For the 16.0 kg cart, gravitational force pulls it down with a force calculated by \( mg \), where \( m \) is mass and \( g \) is gravitational acceleration \( 9.8 \ \mathrm{m/s^2} \).
  • In this exercise, the withstood gravitational force is balanced by the normal force from the ground, allowing no vertical movement.
  • This results in zero work done by gravity, as work requires displacement in its direction of action.
Understanding gravitational force is crucial in such problems, as it confirms assumptions of motion direction and helps establish which forces exceed the pull of gravity.

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 \(1200-\mathrm{kg}\) car is being driven up a \(5.0^{\circ}\) hill. The frictional force is directed opposite to the motion of the car and has a magnitude of \(f=\) \(524 \mathrm{N} .\) A force \(\overrightarrow{\mathbf{F}}\) is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight \(\overrightarrow{\mathbf{W}}\) and the normal force \(\overrightarrow{\mathbf{F}}_{\mathrm{N}}\) directed perpendicular to the road surface. The length of the road up the hill is \(290 \mathrm{m}\). What should be the magnitude of \(\overrightarrow{\mathbf{F}},\) so that the net work done by all the forces acting on the car is \(+150 \mathrm{kJ} ?\)

The brakes of a truck cause it to slow down by applying a retarding force of \(3.0 \times 10^{3} \mathrm{N}\) to the truck over a distance of \(850 \mathrm{m} .\) What is the work done by this force on the truck? Is the work positive or negative? Why?

An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a mass of \(4.5 \times 10^{4} \mathrm{kg},\) and the force causes its speed to change from 7100 to \(5500 \mathrm{m} / \mathrm{s}\) (a) What is the work done by the force? (b) If the asteroid slows down over a distance of \(1.8 \times 10^{6} \mathrm{m},\) determine the magnitude of the force.

A 63-kg skier coasts up a snow-covered hill that makes an angle of \(25^{\circ}\) with the horizontal. The initial speed of the skier is \(6.6 \mathrm{m} / \mathrm{s}\). After coasting \(1.9 \mathrm{m}\) up the slope, the skier has a speed of \(4.4 \mathrm{m} / \mathrm{s}\). (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the drive force is shut off, the snowmobile coasts to a halt. The snowmobile and its rider have a mass of 136 kg. Under the influence of a drive force of \(205 \mathrm{N}\), it is moving at a constant velocity whose magnitude is \(5.50 \mathrm{m} / \mathrm{s}\). The drive force is then shut off. Find (a) the distance in which the snowmobile coasts to a halt and (b) the time required to do so.
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Electrostatics

Read Explanation

Quantum Physics

Read Explanation

Space Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Solid State Physics

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.