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A chain is held on a frictionless table with onefourth of its length hanging over the edge. If the chain has length \(L=28 \mathrm{~cm}\) and mass \(m=0.012 \mathrm{~kg}\), how much work is required to pull the hanging part back onto the table?

Short Answer

Expert verified
The work required is approximately 0.00103 J.

Step by step solution

01

Determine Length of Hanging Part

The total length of the chain is \(L = 28 \text{ cm}\). Since one-fourth of the chain is hanging over the edge, the length of the hanging part is \( \frac{L}{4} = \frac{28}{4} = 7 \text{ cm}\).
02

Derive Formula for Work

The work done to pull the hanging part of the chain back onto the table is given by the formula for the gravitational work: \( W = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (\(9.8 \text{ m/s}^2\)), and \( h \) is the height moved. Since the center of mass of the hanging part is halfway down, \( h = \frac{7}{2} \text{ cm} = 3.5 \text{ cm} = 0.035 \text{ m} \).
03

Calculate Mass of Hanging Part

The mass of the entire chain is \(0.012 \text{ kg}\). Since one-fourth of the chain is hanging, its mass is \( \frac{0.012}{4} = 0.003 \text{ kg}\).
04

Compute Work Done

Substitute the calculated values into the work formula: \( W = 0.003 \times 9.8 \times 0.035 \). Thus, \( W = 0.001029 \text{ J}\) (Joules). Thus, the work required to pull the hanging part back onto the table is approximately \(0.00103 \text{ J}\).

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. Specifically, it is the energy that an object has because of its height above the ground. When you lift an object, you do work against gravity. This work is stored as potential energy.
To calculate gravitational potential energy (\( U \)), we use the formula:
  • \( U = mgh \)
In this formula, \(m\) stands for mass, \(g\) is the gravitational acceleration (approximately 9.8 m/s\(^2\) on Earth), and \(h\) is the height the object is lifted above the ground.
In the chain exercise, the key part is understanding that when a segment of the chain is hanging off the table, it has gravitational potential energy. When you pull it back onto the table, you're increasing its potential energy.
Mass and Length
Mass and length are fundamental properties of physical objects that interrelate in various mechanics problems. Mass is a measure of the amount of matter in an object, influencing how much an object resists a change in motion. Length, in this context, refers to how much of the chain is involved in an interaction.In the exercise, the chain has a total length \(L\) of 28 cm and a mass \(m\) of 0.012 kg. Calculating the properties of a fraction of this whole involves multiplying these quantities by their respective portions. For example, the hanging part of the chain represents a quarter of the total length. Thus, it is necessary to calculate the mass of that specific section to determine how much work is done when changing its position.
  • Hanging Length = \( \frac{L}{4} = 7 \text{ cm} \)
  • Mass of Hanging Part = \( \frac{m}{4} = 0.003 \text{ kg} \)
Understanding both mass and length is crucial for analyzing the problem correctly.
Center of Mass
The center of mass is a critical concept in physics that allows us to simplify complex problems regarding motion and forces. It is the point where the mass of a system is concentrated and behaves as if all the mass were gathered at that point.
When we deal with motion, calculating the center of mass helps determine how forces will cause the system to move. For the chain problem, its center of mass affects how we calculate the work required to lift the hanging part of the chain back onto the table. From the solution, we know:
  • The center of mass of the hanging part is located at its midpoint due to its uniform density.
  • This midpoint for the 7 cm chain segment is 3.5 cm, which is crucial for determining the effective height \(h\) used in calculating work done in raising the chain back.
Thus, the chain's center of mass plays a key role in determining its potential energy.
Frictionless Motion
Frictionless motion refers to an ideal situation where no resistive force opposes the motion of objects. While real-world scenarios often involve friction, considering frictionless conditions simplifies calculations and allows focus on other forces, like gravity, at play. In the chain problem, the frictionless table ensures there's no additional work required to overcome a resistive force when pulling the chain back onto the table. This allows all work done to be directed solely at moving the chain against gravity, elevating it to the top of the table. This concept is important in evaluating how systems move:
  • Without friction, calculations focus solely on gravitational forces and potential energy changes.
  • Problems simplify to pure energy transformations without energy losses through heat or resistance.
Frictionless conditions, therefore, offer a cleaner theoretical framework to analyze and understand fundamental physics concepts contributed by gravitational interactions.

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Most popular questions from this chapter

A \(3.5 \mathrm{~kg}\) block is accelerated from rest by a compressed spring of spring constant \(640 \mathrm{~N} / \mathrm{m}\). The block leaves the spring at the spring's relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction \(\mu_{k}=0.25 .\) The frictional force stops the block in distance \(D=7.8 \mathrm{~m}\). What are (a) the increase in the thermal energy of the block-floor system, (b) the maximum kinetic energy of the block, and (c) the original compression distance of the spring?

At a certain factory, \(300 \mathrm{~kg}\) crates are dropped vertically from a packing machine onto a conveyor belt moving at \(1.20 \mathrm{~m} / \mathrm{s}\) (Fig. \(8-62\) ). (A motor maintains the belt's constant speed.) The coefficient of kinetic friction between the belt and each crate is \(0.400\). After a short time, slipping between the belt and the crate ceases, and the crate then moves along with the belt. For the period of time during which the crate is being brought to rest relative to the belt, calculate, for a coordinate system at rest in the factory, (a) the kinetic energy supplied to the crate, (b) the magnitude of the kinetic frictional force acting on the crate, and (c) the energy supplied by the motor. (d) Explain why answers (a) and (c) differ.

A \(25 \mathrm{~kg}\) bear slides, from rest, \(12 \mathrm{~m}\) down a lodgepole pine tree, moving with a speed of \(5.6 \mathrm{~m} / \mathrm{s}\) just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

A \(70.0 \mathrm{~kg}\) man jumping from a window lands in an elevated fire rescue net \(11.0 \mathrm{~m}\) below the window. He momentarily stops when he has stretched the net by \(1.50 \mathrm{~m}\). Assuming that mechanical energy is conserved during this process and that the net functions like an ideal spring, find the elastic potential energy of the net when it is stretched by \(1.50 \mathrm{~m}\).

A metal tool is sharpened by being held against the rim of a wheel on a grinding machine by a force of \(180 \mathrm{~N}\). The frictional forces between the rim and the tool grind off small pieces of the tool. The wheel has a radius of \(20.0 \mathrm{~cm}\) and rotates at \(2.50\) rev/s. The coefficient of kinetic friction between the wheel and the tool is \(0.320 .\) At what rate is energy being transferred from the motor driving the wheel to the thermal energy of the wheel and tool and to the kinetic energy of the material thrown from the tool?

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