/*! 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 28 A 75 -kg rooter climbs a vertica... [FREE SOLUTION] | 91Ó°ÊÓ

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A 75 -kg rooter climbs a vertical 7.0 -m ladder to the flat roof of a house. He then walks 12 \(\mathrm{m}\) on the roof, climbs down another vertical 7.0 -m ladder, and finally walks on the ground back to his starting point. How much work is done on him by gravity (a) as he climbs up; (b) as he climbs down; (c) as he walks on the roof and on the ground? (d) What is the total work done on him by gravity during this round trip? (e) On the basis of your answer to part (d), would you say that gravity is a conservative or nonconservative force? Explain.

Short Answer

Expert verified
(d) Total work is 0 J. (e) Gravity is a conservative force.

Step by step solution

01

Understanding Work Done by Gravity

Work done by gravity is calculated using the formula \( W = mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity (approximated as \( 9.8 \, \mathrm{m/s^2} \)), and \( h \) is height. The gravity force acts vertically.
02

Work Done as He Climbs Up

As the rooter climbs a 7.0 m vertical ladder, the work done by gravity is calculated as: \( W_{up} = mgh = 75 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 7.0 \, \mathrm{m} \). Evaluating gives \( W_{up} = 5145 \, \mathrm{J} \). Since the force and displacement are opposite, this work is negative.
03

Work Done as He Climbs Down

As he climbs down the 7.0 m ladder, the gravity assists him, so the work is positive. Calculate: \( W_{down} = mgh = 75 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 7.0 \, \mathrm{m} \). This results in \( W_{down} = 5145 \, \mathrm{J} \).
04

Work Done While Walking

Walking on the roof or the ground doesn't result in work done by gravity because there is no vertical displacement. Hence, \( W_{walk} = 0 \).
05

Total Work Done by Gravity During Round Trip

The total work done by gravity is the sum of work while climbing up, down, and walking. \( W_{total} = W_{up} + W_{down} + W_{walk} = -5145 \, \mathrm{J} + 5145 \, \mathrm{J} + 0 = 0 \, \mathrm{J} \).
06

Nature of Gravity as a Force

Since the total work done by gravity is zero for the entire round trip, gravity is considered a conservative force. Conservative forces have path-independent work and depend only on initial and final positions.

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Key Concepts

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

Understanding Conservative Forces
In the realm of physics, conservative forces are an essential concept that students need to grasp. These forces, such as gravity, have a unique characteristic: the work done by them is independent of the path taken. Imagine a ball rolling up and down a hill. No matter how it moves, the work done by gravity depends only on the height difference between the start and end points.
Conservative forces have a key property that the work done around a closed loop is zero. Such forces also allow the definition of potential energy, a stored form of energy arising from an object's position. This is because the work done is recoverable without loss. In the context of our exercise, we find that the total work done by gravity during the roofer’s round trip is zero, confirming gravity's nature as a conservative force.
  • Path independence: Work depends on initial and final positions.
  • Closed loop work: Total is zero for any closed path.
  • Potential energy: Can be defined and recovered without loss.
Simplifying Physics Problem Solving
Physics problem-solving can sometimes seem complex, but breaking it down into steps makes it manageable. For problems involving forces like gravity, it's crucial to understand and apply the basic principles and formulas.
The exercise can be approached by following a structured process:
  • Identify the forces at play: In this case, gravity.
  • Determine the work done: Use the formula for work, which involves mass, gravitational acceleration, and height.
  • Break the problem into parts: Calculate work for each stage - climbing up, climbing down, and walking on flat surfaces.
  • Solve step-by-step: Carefully compute each part before summing them up.
  • Analyze the results: Interpret the meaning of zero total work as it relates to the properties of force.
These methods not only make solving such problems easier but also deepen understanding of physics concepts. Practicing structured problem-solving enhances both speed and accuracy in addressing related exercises.
Gravitational Potential Energy Explained
Gravitational potential energy, often denoted as GPE, is a form of energy that relates to an object's height above a reference point. When an object is lifted against gravity, it gains energy based on its mass, the gravitational pull, and the height it ascends.
The formula for gravitational potential energy is: \[ U = mgh \]where:\
  • - \( U \) is the gravitational potential energy,
    \
  • - \( m \) 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 above the reference level.
    \In our scenario, when the roofer climbs up the ladder, he accumulates potential energy. This energy decreases when he descends. On flat ground or the roof top, his potential energy doesn't change, as there's no change in height. Thus, no work by gravity occurs there.
    Recognizing the role of gravitational potential energy helps us understand why gravity is a conservative force - as the man's total energy related to gravity is constant if height changes are cyclical.
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    Most popular questions from this chapter

    EALE A force parallel to the \(x\) -axis acts on a particle moving along the \(x\) -axis. This force produces potential energy \(U(x)\) given by \(U(x)=\alpha x^{4},\) where \(\alpha=1.20 \mathrm{J} / \mathrm{m}^{4} .\) What is the force (magnitude and direction) when the particle is at \(x=-0.800 \mathrm{m}\) ?

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