Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

How much work do you do when you hold a bag of groceries while standing still? How much work do you do when carrying the same bag a distance \(d\) across the parking lot of the grocery store?

Short Answer

Expert verified
Question: Calculate the work done when holding a bag of groceries and standing still, and when carrying it a distance d across the parking lot. Answer: In both situations, no work is done. When standing still, there is no displacement, and when carrying the bag horizontally, the force and direction of movement are perpendicular, resulting in a work of 0 joules in both cases.

Step by step solution

01

When holding the bag of groceries while standing still:

In this scenario, there is no displacement of the grocery bag, which means that the distance moved is zero (\(d=0\)). Therefore, the work done while holding the grocery bag and standing still is: \(W=F \cdot 0 \cdot \cos{\theta} = 0\) joules. No work is done when holding the bag of groceries and standing still.
02

When carrying the bag of groceries a distance \(d\) across the parking lot:

In this scenario, the bag of groceries is moved horizontally for a distance \(d\). The force holding the groceries still acts in the vertical direction, and since the movement is horizontal, the angle \(\theta\) is \(90^\circ\). Now we can apply the formula for work: \(W = Fd\cos{90^\circ}\) Since \(\cos{90^\circ} = 0\): \(W = Fd \cdot 0 = 0\) joules. No work is done when carrying the bag of groceries a distance \(d\) across the parking lot, as the force applied and the direction of movement are perpendicular to each other.

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 Vaia!

Key Concepts

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

Work-Energy Theorem
The work-energy theorem is a fundamental principle in physics that connects the concepts of work and energy. The theorem states that the work done by all forces acting on an object is equal to the change in its kinetic energy. This concept is written mathematically as:
\[\begin{equation}W = \triangle KE = KE_{final} - KE_{initial}\right)\text{where}\end{equation}\]
\( W \) is the work done, \( \triangle KE \) is the change in kinetic energy, \( KE_{final} \) is the kinetic energy of the object at the end, and \( KE_{initial} \) is the kinetic energy at the beginning. In basic terms, if you push or pull on an object causing it to move, you have done work on it, and this work results in a change in the object's kinetic energy. However, if there is no movement or if the force does not cause displacement, like holding a bag without moving it, no work is transferred, and the kinetic energy of the bag remains unchanged. This explains why no work is done when one stands still holding a bag or when the force applied to carry the bag does not result in its displacement in the direction of the force.
Displacement in Physics
Displacement in physics is defined as the change in the position of an object. It is a vector quantity, which means it possesses both magnitude and direction. Displacement is represented mathematically as the final position minus the initial position:
\[\begin{equation}\vec{d} = \vec{x}_{final} - \vec{x}_{initial}\right)\text{where}\end{equation}\]
\( \vec{d} \) is the displacement vector, and \( \vec{x}_{final} \) and \( \vec{x}_{initial} \) are the final and initial position vectors, respectively. The importance of displacement lies in its role in the definition of work done by a force. To calculate work, one must take into account the displacement of the object in the direction of the force. If there is no displacement, as in the case of holding a bag stationary, the displacement is zero, and hence, no work is performed, regardless of the amount of force applied. Conversely, even if the bag is carried across a parking lot, if the force applied is perpendicular to the displacement (in this case, holding the bag while walking), displacement in the direction of the force is still zero, resulting in no work done in the context of physics.
Force and Work
Force and work in physics are closely related, yet distinct concepts. A force is a push or pull on an object that can cause it to accelerate, while work occurs when a force causes an object to move through a displacement. The relationship between force and work is expressed by the equation:
\[\begin{equation}W = F \times d \times \begin{document}\cos(\theta)\right)\text{where}\end{document}
\( W \) is work, \( F \) is the magnitude of the force applied, \( d \) is the magnitude of the displacement, and \( \theta \) is the angle between the force vector and the displacement vector. When the force is parallel to the displacement, maximum work is done; when it is perpendicular, no work is done. This concept is exemplified in the grocery bag example: while you may exert a force to hold the bag, if the bag does not move, or if you carry the bag such that the displacement is perpendicular to the force of gravity, then the work done is zero. Understanding the necessity of both a force and a resultant displacement in the same direction is key to comprehending when work is performed in a physical sense.

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 rock of mass \(0.773 \mathrm{~kg}\) is hanging from a string of length \(2.45 \mathrm{~m}\) on the Moon, where the gravitational acceleration is a sixth of that on Earth. What is the change in gravitational potential energy of this rock when it is moved so that the angle of the string changes from \(3.31^{\circ}\) to \(14.01^{\circ}\) ? (Both angles are measured relative to the vertical.)

A child throws three identical marbles from the same height above the ground so that they land on the flat roof of a building. The marbles are launched with the same initial speed. The first marble, marble \(\mathrm{A}\), is thrown at an angle of \(75^{\circ}\) above horizontal, while marbles \(B\) and \(C\) are thrown with launch angles of \(60^{\circ}\) and \(45^{\circ}\), respectively. Neglecting air resistance, rank the marbles according to the speeds with which they hit the roof. a) \(A

A \(1.00-\mathrm{kg}\) block compresses a spring for which \(k=100 . \mathrm{N} / \mathrm{m}\) by \(20.0 \mathrm{~cm} ;\) the spring is then released, and the block moves across a horizontal, frictionless table, where it hits and compresses another spring, for which \(k=50.0 \mathrm{~N} / \mathrm{m} .\) Determine a) the total mechanical energy of the system, b) the speed of the mass while moving freely between springs, and c) the maximum compression of the second spring.

Can a unique potential energy function be identified with a particular conservative force?

A 0.500 -kg mass is attached to a horizontal spring with \(k=100 . \mathrm{N} / \mathrm{m}\) The mass slides across a frictionless surface. The spring is stretched \(25.0 \mathrm{~cm}\) from equilibrium, and then the mass is released from rest. a) Find the mechanical energy of the system. b) Find the speed of the mass when it has moved \(5.00 \mathrm{~cm}\). c) Find the maximum speed of the mass.

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.

Sign-up for free