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Chapter 8: Problem 37

At a lunar base, a uniform chain hangs over the edge of a horizontal platform. A machine does \(1.0 \mathrm{~J}\) of work in pulling the rest of the chain onto the platform. The chain has a mass of \(2.0\) \(\mathrm{kg}\) and a length of \(3.0 \mathrm{~m}\). What length was initially hanging over the edge? On the Moon, the gravitational acceleration is \(1 / 6\) of \(9.8 \mathrm{~m} / \mathrm{s}^{2}\).

Short Answer

Expert verified
The length of the chain initially hanging over the edge was approximately 1.36 meters.

Step by step solution

01

Understand the Problem

The problem involves a chain hanging over the edge of a platform on the Moon. We need to determine the length of the chain that was initially hanging. We're given the total work done to pull the chain, the chain's mass and length, and the Moon's gravitational acceleration, which is \( \frac{1}{6} \times 9.8 \, \mathrm{m/s^2} \).
02

Calculate the Gravitational Force

First, calculate the gravitational force per unit length of the chain. The gravitational force on the Moon is \( g = \frac{9.8}{6} \, \mathrm{m/s^2} \). The force per unit length \( F \) of the chain is calculated by \( F = \text{mass per unit length} \times g = \frac{2.0 \, \mathrm{kg}}{3.0 \, \mathrm{m}} \times 1.63 \, \mathrm{m/s^2} \). This gives a force per unit length of approximately \( 1.0867 \, \mathrm{N/m} \).
03

Relate Work Done to Chain Length

The work done by the machine to pull the chain back onto the platform is effectively against gravity. If \( x \) is the length hanging over the edge, the work done is \( W = F \times x \times \frac{x}{2} = 1.0 \, \mathrm{J} \), where \( \frac{x}{2} \) is the average height (center of mass) of the hanging chain.
04

Solve for the Length of the Chain Hanging

Substitute the gravitational force per unit length into the work formula: \[ 1.0 = 1.0867 \times \frac{x^2}{2} \]Rearrange and solve for \( x^2 \) to find \( x \).\[ x^2 = \frac{2.0}{1.0867} \approx 1.84 \]Taking the square root of both sides gives:\[ x \approx \sqrt{1.84} \approx 1.36 \, \mathrm{m} \]
05

Verify the Solution

Check if the calculations make sense. Substitute \( x = 1.36 \, \mathrm{m} \) back into the work equation to confirm the amount of work done is approximately \( 1.0 \, \mathrm{J} \). Since the calculated work closely matches the given work, the solution is verified.

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Key Concepts

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

Gravitational Force
Understanding how gravitational force operates is key to solving many physics problems, especially those involving celestial bodies like the Moon. Gravitational force is the attractive force that a massive object exerts on another. On Earth, this force accelerates objects at approximately \( 9.8 \, \text{m/s}^2 \). On the Moon, however, the gravitational acceleration is much weaker. In our problem, gravity on the Moon is \( \frac{1}{6} \) of Earth's, making it \( 1.63 \, \text{m/s}^2 \). This reduced gravity affects how we calculate the forces acting on the chain.
For practical purposes, when dealing with a chain, we need to calculate the gravitational force per unit length of the chain. This helps simplify the problem as it distributes the total force along the length of the chain, allowing us to focus on the work needed to lift individual portions. In simple terms, each meter of the chain experiences a downward pull due to gravity which we calculated as approximately \( 1.0867 \, \text{N/m} \). Understanding this is crucial for effectively calculating how much work is required to move the chain.
Work and Energy
Work and energy concepts play a central role in this problem. Work is defined in physics as the amount of energy transferred by a force over a distance. The formula for calculating work is \( W = F \times d \), where \( F \) is the force applied, and \( d \) is the distance over which the force is applied. When pulling the chain onto the platform, the machine exerts a force over the length of the chain that hangs off the edge.
Since the chain is uniform, the force exerted by the machine does more work on parts of the chain that are further from the edge than on parts closer. Essentially, this force combats the gravitational force. When considering this, it is helpful to think about the process in segments: each section of the chain being slightly more work-intensive to lift than the one before. In our exercise, we calculated that the total work performed by the machine was \( 1.0 \, \text{J} \). This is the amount of energy expended to lift the center of mass of the hanging part of the chain.
Center of Mass
The concept of the center of mass is integral in understanding how objects behave under various forces. For a uniform object, like our chain, the center of mass is located at its midpoint. When the chain hangs over the edge, the center of mass of the hanging portion is crucial because it indicates where the cumulative effect of gravitational forces is applied.
In our example, the chain initially hangs over the platform, and its center of mass determines how the gravitational force is distributed. As the chain is lifted, the portion remaining on the platform increases, shifting the center of mass. The work done by the machine involves lifting this average position, not the entire length at once. This simplification is useful in calculating dynamics, as it focuses on where the effective "center" of the weight is located.
  • The center of mass is midway through the hanging part.
  • It guides how gravitational forces are distributed.
  • Understanding its role simplifies calculations significantly.

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Most popular questions from this chapter

A loaded truck of mass \(3000 \mathrm{~kg}\) moves on a level road at a constant speed of \(6.000 \mathrm{~m} / \mathrm{s}\). The frictional force on the truck from the road is \(1000 \mathrm{~N}\). Assume that air drag is negligible. (a) How much work is done by the truck engine in \(10.00 \mathrm{~min}\) ? (b) After \(10.00 \mathrm{~min}\), the truck enters a hilly region whose inclination is \(30^{\circ}\) and continues to move with the same speed for another \(10.00 \mathrm{~min}\). What is the total work done by the engine during that period against the gravitational force and the frictional force? (c) What is the total work done by the engine in the full \(20 \mathrm{~min}\) ?

A \(10.0 \mathrm{~kg}\) block falls \(30.0 \mathrm{~m}\) onto a vertical spring whose lower end is fixed to a platform. When the spring reaches its maximum compression of \(0.200 \mathrm{~m}\), it is locked in place. The block is then removed and the spring apparatus is transported to the Moon, where the gravitational acceleration is \(\mathrm{g} / 6 . \mathrm{A} 50.0 \mathrm{~kg}\) astronaut then sits on top of the spring and the spring is unlocked so that it propels the astronaut upward. How high above that initial point does the astronaut rise?

A conservative force \(\vec{F}=(6.0 x-12) \hat{i} \mathrm{~N}\), where \(x\) is in meters, acts on a particle moving along an \(x\) axis. The potential energy \(U\) associated with this force is assigned a value of \(27 \mathrm{~J}\) at \(x=0\). (a) Write an expression for \(U\) as a function of \(x\), with \(U\) in joules and \(x\) in meters. (b) What is the maximum positive potential energy? At what (c) negative value and (d) positive value of \(x\) is the potential energy equal to zero?

A \(4.0 \mathrm{~kg}\) bundle starts up a \(30^{\circ}\) incline with \(150 \mathrm{~J}\) of kinetic energy. How far will it slide up the incline if the coefficient of kinetic friction between bundle and incline is \(0.36 ?\)

A horizontal force of magnitude \(41.0 \mathrm{~N}\) pushes a block of mass \(4.00 \mathrm{~kg}\) across a floor where the coefficient of kinetic friction is \(0.600\). (a) How much work is done by that applied force on the block-floor system when the block slides through a displacement of \(2.00 \mathrm{~m}\) across the floor? (b) During that displacement, the thermal energy of the block increases by \(40.0 \mathrm{~J}\). What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block?
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