/*! 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 8 You pick up a \(3.4-\mathrm{kg}\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You pick up a \(3.4-\mathrm{kg}\) can of paint from the ground and lift it to a height of \(1.8 \mathrm{m}\). (a) How much work do you do on the can of paint? (b) You hold the can stationary for half a minute, waiting for a friend on a ladder to take it. How much work do you do during this time? (c) Your friend decides against the paint, so you lower it back to the ground. How much work do you do on the can as you lower it?

Short Answer

Expert verified
(a) 59.976 J, (b) 0 J, (c) -59.976 J.

Step by step solution

01

Understand the Work Formula

The work done on an object against gravity is calculated using the formula: \( W = mgh \), where \( W \) is the work done, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( h \) is the height.
02

Calculate Work While Lifting

Substitute the values into the formula to calculate the work done while lifting the can. Here, \( m = 3.4 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( h = 1.8 \, \text{m} \). Thus, \( W = 3.4 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1.8 \, \text{m} \). Calculate this to find \( W \).
03

Calculate Force

For part (a), calculate using \( W = 3.4 \times 9.8 \times 1.8 \). Perform the multiplication step by step: \( 3.4 \times 9.8 = 33.32 \), then \( 33.32 \times 1.8 = 59.976 \). Thus, the work done is \( 59.976 \, \text{J} \).
04

Determine Work While Holding

When you hold the can stationary, no work is being done on the can, because work requires movement. Thus, the work done is \( 0 \, \text{J} \).
05

Calculate Work While Lowering

When lowering the can, the same amount of work is done in magnitude but in the opposite direction (downward). So, the work done is \(-59.976 \, \text{J} \) because you are exerting force opposite to the direction of motion.

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.

Gravitational Potential Energy
Gravitational Potential Energy (GPE) is the energy an object possesses due to its position in a gravitational field. When you lift an object, like a can of paint, you do work against the force of gravity. This work then gets stored as gravitational potential energy.- The key formula for GPE is given by \( E_p = mgh \) where: - \( m \) is the mass in kilograms - \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)) - \( h \) is the height in meters above the reference levelWhen you lifted the can of paint, you increased its gravitational potential energy by doing work on it. Similarly, when you lower it back to the ground, the energy is released back as work.
Mass
Mass is a fundamental property of physical objects and is a measure of the amount of matter in an object. It does not change regardless of the object's location in the universe. - In the context of physics problems involving gravity, the mass is a critical factor in calculating: - Gravitational force - Work done against or by gravity Leverage mass in calculations as it directly influences the work done when lifting or lowering an object. For example, in the exercise, the can's mass of 3.4 kg was crucial in determining how much work you did when lifting it against earth’s gravity.
Newton's Laws
Newton's Laws of Motion help explain the forces and interactions of moving objects. In this problem, Newton's Second Law, which states that force equals mass times acceleration (\( F = ma \)), is particularly relevant.- Let's break it down: - Lifting the can involves applying a force to overcome gravitational pull. - While holding the can stationary, there's no displacement, hence no work is done.The work done during lifting implies you are applying a constant upward force (equal to the gravitational force on the can) over a distance. This relationship aligns with Newton's Laws and gives a clear picture of the required effort to lift an object.
Physics Problems
Physics problems often use real-life scenarios to apply theoretical concepts such as work, energy, and force. The paint can exercise is a practical introduction to work, showcasing how theoretical physics principles manifest in everyday activities. - Essentials when solving physics problems: - Understand the relationship between forces and movements - Apply principles like conservation of energy - Use mathematical formulas to quantify processes By practicing these applications, you can more clearly see the role that physics plays in both simple and complex situations, sharpening your problem-solving skills.

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

Children in a tree house lift a small dog in a basket \(4.70 \mathrm{m}\) up to their house. If it takes 201 J of work to do this, what is the combined mass of the dog and basket?

IP A \(1300-\mathrm{kg}\) car delivers a constant 49 hp to the drive wheels. We assume the car is traveling on a level road and that all frictional forces may be ignored. (a) What is the acceleration of this car when its speed is \(14 \mathrm{m} / \mathrm{s} ?\) (b) If the speed of the car is doubled, does its acceleration increase, decrease, or stay the same? Explain. (c) Calculate the car's acceleration when its speed is \(28 \mathrm{m} / \mathrm{s}\).

BIO Climbing the Empire State Building A new record for running the stairs of the Empire State Building was set on February \(3,2003 .\) The 86 flights, with a total of 1576 steps, was run in 9 minutes and 33 seconds. If the height gain of each step was \(0.20 \mathrm{m},\) and the mass of the runner was \(70.0 \mathrm{kg}\), what was his average power output during the climb? Give your answer in both watts and horsepower.

IP A grandfather clock is powered by the descent of a \(4.35-\mathrm{kg}\) weight. (a) If the weight descends through a distance of \(0.760 \mathrm{m}\) in 3.25 days, how much power does it deliver to the clock? (b) To increase the power delivered to the clock, should the time it takes for the mass to descend be increased or decreased? Explain.

To keep her dog from running away while she talks to a friend, Susan pulls gently on the dog's leash with a constant force given by \(\overrightarrow{\mathbf{F}}=(2.2 \mathrm{N}) \hat{\mathrm{x}}+(1.1 \mathrm{N}) \hat{\mathrm{y}} .\) How much work does she do on the dog if its displacement is (a) \(\overrightarrow{\mathrm{d}}=(0.25 \mathrm{m}) \hat{\mathrm{x}}\) (b) \(\overrightarrow{\mathrm{d}}=(0.25 \mathrm{m}) \hat{\mathrm{y}}, \mathrm{or}\) (c) \(\overrightarrow{\mathrm{d}}=(-0.50 \mathrm{m}) \hat{\mathrm{x}}+(-0.25 \mathrm{m}) \hat{\mathrm{y}} ?\)

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.