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Chapter 6: Problem 66

A uniform chain of length \(2 \mathrm{~m}\) is kept on a table such that a length of \(60 \mathrm{~cm}\) hangs freely from the edge of the table. The total mass of the chain is \(4 \mathrm{~kg}\). What is the work done in pulling the entire chain on the table? Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) [2004] (A) \(12 \mathrm{~J}\) (B) \(3.6 \mathrm{~J}\) (C) \(7.2 \mathrm{~J}\) (D) \(1200 \mathrm{~J}\)

Short Answer

Expert verified
The work done in pulling the entire chain on the table is \(3.6 \mathrm{~J}\).

Step by step solution

01

Determine the mass per unit length of the chain

Since the total mass of the chain is given as 4 kg and its length is 2 meters, we can find the mass per unit length by dividing the mass by the length. In this case, we have: Mass per unit length = \(\frac{4 \mathrm{~kg}}{2 \mathrm{~m}} = 2 \frac{\mathrm{kg}}{\mathrm{m}}\)
02

Calculate the initial potential energy

Now we must determine the initial potential energy of the hanging chain, which will help us find the difference in potential energy when all the chain is pulled onto the table. We know that part of the chain hanging off the table is 60 cm or 0.6 m. Since the mass per unit length is 2 kg/m, the mass of hanging part is: \(m_{hanging} = 2 \frac{\mathrm{kg}}{\mathrm{m}} \cdot 0.6 \mathrm{~m} = 1.2 \mathrm{~kg}\) We can now compute the initial potential energy: \(PE_{initial} = m_{hanging} \cdot g \cdot \frac{0.6 \mathrm{~m}}{2}\) \(PE_{initial} = 1.2 \mathrm{~kg} \cdot 10 \frac{\mathrm{m}}{\mathrm{s}^{2}} \cdot 0.3 \mathrm{~m} = 3.6 \mathrm{~J}\)
03

Calculate the final potential energy

When all the chain is pulled onto the table, the potential energy of the hanging section will be zero, because there will be no height difference. So, \(PE_{final}=0\)
04

Calculate the work done

The work done is the difference in potential energy between the initial and final states. So, Work done = \(PE_{final} - PE_{initial}\) Work done = \(0 - 3.6 \mathrm{~J} = -3.6 \mathrm{~J}\) Since the negative sign represents the work is done against the gravitational force (it means work is positive for acting force), we can take the absolute value of work done: Work done = |(-3.6 \mathrm{~J})| = \(3.6 \mathrm{~J}\) This corresponds to answer choice (B).

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Key Concepts

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

Potential Energy
Potential energy is the stored energy an object possesses due to its position or state. For example, when part of a chain hangs from a table, as in the problem, the portion over the edge has gravitational potential energy because of its height above the ground.

This energy is calculated using the formula:
\[ PE = m \times g \times h \]
where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approx. \( 10 \frac{m}{s^2} \) on Earth), and \( h \) is the height of the object's center of mass above the reference point. For the hanging chain, the potential energy decreases to zero once it is entirely on the table, signifying a transfer of energy as work to pull it up.

In understanding physics work done problems, realizing that work is a transfer of energy is crucial. The work done to lift the chain is equal to the loss of its potential energy.
Mass per Unit Length
Mass per unit length is a measure of how the mass of an object is distributed along its length. In our chain example, it's calculated by dividing the total mass by the total length of the chain. This value is essential because it helps determine the mass of a segment of the chain without needing to measure it separately.

Here鈥檚 the formula used:\[ \text{Mass per unit length} = \frac{\text{Total mass}}{\text{Total length}} \]
Knowing this concept allows us to calculate the mass of the chain section that hangs off the table. When dealing with non-uniform objects, the mass per unit length might change along the length, but for uniform objects like the chain in our problem, it remains constant, simplifying calculations of potential energy.
Gravitational Force
Gravitational force is the attraction between two bodies due to their masses. In physics, it's a fundamental interaction that explains how objects are attracted towards Earth, giving them weight.

The force of gravity on an object near Earth's surface is often expressed as:\[ F = m \times g \]
Where \( F \) is the gravitational force, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity. This force is what causes the potential energy of the hanging chain to exist. When work is done to lift the chain, it's overcoming the gravitational force pulling it down. The work done is essentially transferring energy to change the chain's height and, consequently, its potential energy.
Gravitational force is central to many physics work done problems because it often provides the resistance that must be overcome by work, as demonstrated in the problem where we calculate the work done to lift the chain onto the table.

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Most popular questions from this chapter

For a particle in uniform circular motion, the acceleration \(\vec{a}\) at a point \(P(R, \theta)\) on the circle of radius \(R\) is (here \(\theta\) is measured from the \(x\)-axis) [2010](A) \(-\frac{v^{2}}{R} \cos \theta \hat{i}+\frac{v^{2}}{R} \sin \theta \hat{j}\) (B) \(-\frac{v^{2}}{R} \sin \theta \hat{i}+\frac{v^{2}}{R} \cos \theta \hat{j}\) (C) \(-\frac{v^{2}}{R} \cos \theta \hat{i}-\frac{v^{2}}{R} \sin \theta \hat{j}\) (D) \(\frac{v^{2}}{R} \hat{i}+\frac{v^{2}}{R} \hat{j}\)

A mass \(m\) of radius \(r\) is rolling horizontally without any slip with a linear speed \(v .\) It then rolls up to a height given by \(\frac{3}{4} \frac{v^{2}}{g}\) (A) the body is identified to be a disc or a solid cylinder. (B) the body is a solid sphere. (C) MOI of the body about instantaneous axis of rotation is \(\frac{1}{2} m r^{2}\). (D) MOI of the body about instantaneous axis of rotation is \(\frac{2}{5} m r^{2}\).

A string is wrapped several times round a solid cylinder of mass \(m\) and then the end of the string is held stationary while the cylinder is released from rest with no initial motion. The acceleration of the cylinder will be \(\frac{n g}{3}\), then the value of \(n\) is.

A mass \(m\) is supported by a massless string wound around a uniform hollow cylinder of mass \(m\) and radius \(R\). If the string does not slip on the cylinder, with what acceleration will the mass fall on release? (A) \(\frac{2 g}{3}\) (B) \(\frac{g}{2}\) (C) \(\frac{5 g}{6}\) (D) \(\mathrm{g}\)

While the disc skids, its translational and angular acceleration are related as (A) \(a=R \alpha\) (B) \(a=\frac{1}{4} R \alpha\) (C) \(a=\frac{1}{2} R \alpha\) (D) \(a=2 R \alpha\)
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