/*! 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 6\. A person pushes a 16.0-kg sh... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(48.0-N\) 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
(a) 54.9 N, (b) 1050 J, (c) -1056 J, (d) 0 J.

Step by step solution

01

Determine the Required Force

To push the cart at a constant velocity, the net force must be zero. This means the horizontal component of the force applied by the shopper, \( F_{x} \), must balance the frictional force of 48 N. The force exerted by the shopper, \( F \), can be found using trigonometry. Calculate the horizontal component as follows:\[F_{x} = F \cdot \cos(29^{\circ}) = 48 \, \text{N}\]Solve for \( F \):\[F = \frac{48}{\cos(29^{\circ})}\]Plugging in the values, we find:\[F \approx 54.9 \, \text{N}\]
02

Calculate Work Done by Pushing Force

The work done by the pushing force is computed using the formula:\[W = F \cdot d \cdot \cos(\theta)\]where \( F = 54.9 \, \text{N} \), \( d = 22.0 \, \text{m} \), and \( \theta = 29^{\circ} \). Plug in these values:\[W = 54.9 \, \text{N} \times 22.0 \, \text{m} \times \cos(29^{\circ})\]This yields:\[W \approx 1050 \, \text{J}\]
03

Calculate Work Done by Frictional Force

The work done by friction can be calculated using the formula:\[W_{ ext{friction}} = F_{ ext{friction}} \times d \times \cos(180^{\circ})\]Compute it with \( F_{ ext{friction}} = 48 \, \text{N} \), \( d = 22.0 \, \text{m} \). Since the frictional force is opposite to the direction of motion, we use the angle \( 180^{\circ} \):\[W_{ ext{friction}} = 48 \, \text{N} \times 22.0 \, \text{m} \times (-1)\]This gives:\[W_{ ext{friction}} = -1056 \, \text{J}\]
04

Calculate Work Done by Gravitational Force

The gravitational force acts vertically and has no displacement in the direction of the force, thus:\[W_{ ext{gravity}} = F_{g} \cdot d \cdot \cos(90^{\circ})\]Since \( \cos(90^{\circ}) = 0 \):\[W_{ ext{gravity}} = 0 \, \text{J}\]
05

Final Step: Review and Summarize the Answers

From the above calculations, we gather the following results: - The magnitude of the shopper's force is approximately 54.9 N. - The work done by the pushing force is approximately 1050 J. - The work done by the frictional force is -1056 J. - The work done by the gravitational force is 0 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 91Ó°ÊÓ!

Key Concepts

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

Understanding Work in Physics
In physics, work is a concept that involves a force causing an object to move. Work is calculated using the formula:
  • \[W = F \cdot d \cdot \cos(\theta)\]
Here, \(W\) represents work, \(F\) is the force applied, \(d\) is the distance over which the force acts, and \(\theta\) is the angle between the force and the direction of motion.
This equation helps us determine how much energy is transferred when an object moves under the influence of a force.
It's crucial to understand that work is only done when the movement of an object aligns with the direction of the force applied. For instance, if the force is perpendicular to the motion, like in the case of gravitational force on a horizontal movement, no work is done. Work can be positive when the force's direction aligns with the displacement and negative if opposite, such as with friction.
Exploring the Concept of Force
Force is a push or pull acting on an object, resulting from its interaction with another object. When multiple forces act on an object, the net force determines its motion. In the context of our shopping cart exercise, the shopper's force is crucial to balance out friction and other forces.

We use trigonometry to resolve the direction and magnitude of forces. The force exerted by the shopper is calculated using trigonometric functions:
  • The horizontal component is \( F_{x} = F \cdot \cos(29^{\circ}) \).
This component must counteract the frictional force to maintain constant velocity.
Thus, solving for \(F\) involves understanding this relationship and using trigonometry to ensure the proper balancing of forces in the horizontal plane.
The Role of Friction in Motion
Friction is the force that opposes the relative motion or tendency of such motion between two surfaces in contact. It acts opposite to the direction of motion and can make it more challenging to move objects, like the shopping cart in the exercise. The shopping cart is subject to a 48 N frictional force.
The work done by friction is always negative since its force direction opposes displacement. This is calculated using:
  • \[W_{friction} = F_{friction} \times d \times \cos(180^{\circ})\]
Here, the use of \(\cos(180^{\circ})\) reflects the opposing nature, leading to negative work. This negative value indicates energy is being used to overcome friction.
Using Trigonometry in Force Analysis
Trigonometry plays a fundamental role in breaking down forces into components, especially when forces act at angles. For the shopper pushing the cart at 29 degrees below the horizontal, we employ trigonometric functions to resolve the force accurately.
In our calculations:
  • We use \(\cos(\theta)\) to determine the horizontal force component.
  • This allows us to relate the applied force to the necessary counterpart that balances friction.
Trigonometry enables precise calculations in physics by addressing the angled force applications and ensuring each component of force is correctly applied in analyses, maintaining clarity and accuracy in problem-solving scenarios.

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 person is making homemade ice cream. She exerts a force of magnitude \(22 \mathrm{N}\) on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius \(0.28 \mathrm{m}\). The force is always applied parallel to the motion of the handle. If the handle is turned once every \(1.3 \mathrm{s},\) what is the average power being expended?

A 75.0-kg skier rides a 2830-m-long lift to the top of a mountain. The lift makes an angle of \(14.6^{\circ}\) with the horizontal. What is the change in the skier's gravitational potential energy?

A car accelerates uniformly from rest to \(20.0 \mathrm{m} / \mathrm{s}\) in \(5.6 \mathrm{s}\) along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is \(9.0 \times 10^{3} \mathrm{N}\) and (b) the weight of the car is \(1.4 \times 10^{4} \mathrm{N}\).

You are working out on a rowing machine. Each time youpull the rowing bar (which simulates the oars) toward you, it moves a distance of \(1.2 \mathrm{m}\) in a time of \(1.5 \mathrm{s}\). The readout on the display indicates that the average power you are producing is 82 W. What is the magnitude of the force that you exert on the handle?

The drawing shows a skateboarder moving at \(5.4 \mathrm{m} / \mathrm{s}\) along a horizontal section of a track that is slanted upward by \(48^{\circ}\) above the horizontal at its end, which is \(0.40 \mathrm{m}\) above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height \(H\) to which she rises above the end of the track.

See all solutions

Recommended explanations on Physics Textbooks

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.