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Chapter 10: Problem 35

Chain on Table a chain is held on a frictionless table with one-fourth of its length hanging over the edge. If the chain has length \(L\) and mass \(m\), how much work is required to pull the hanging part back onto the table?

Short Answer

Expert verified
\(\frac{m g L}{32}\)

Step by step solution

01

- Understand the Problem

We need to calculate the work required to pull one-fourth of a chain (with length L and mass m) back onto a frictionless table. The hanging part will be pulled back onto the table, working against gravity.
02

- Determine the Mass of the Hanging Part

First, calculate the mass of the hanging part of the chain. Since one-fourth of the chain is hanging, the mass of the hanging part is \(\frac{m}{4}\).
03

- Set Up the Work Integral

The work needed to pull the mass onto the table is equivalent to the work against gravity. The distance through which the chain鈥檚 center of mass (for the hanging part) is pulled is \(\frac{L}{8}\), since the center of mass of the hanging segment is halfway along its length.
04

- Calculate the Work Done to Pull the Chain

Use the formula for work against gravity: \(\text{Work} = \text{Force} \times \text{Distance}\). Here, the force is the weight of the hanging part (\frac{m}{4} \times g) and the distance is \(\frac{L}{8}\). So, \(\text{Work} = (\frac{m}{4} \times g) \times \frac{L}{8}\).
05

- Simplify the Expression

Simplify the work done: \(\text{Work} = \frac{m \times g \times L}{32}\). Hence, the total work required to pull the hanging part of the chain back onto the table is \(\frac{m g L}{32}\).

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Key Concepts

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

work against gravity
In physics, work against gravity refers to the energy required to move an object against the force of gravity. Essentially, it鈥檚 about overcoming the weight of the object to move it vertically.
In this specific problem, we need to pull a part of the chain onto a frictionless table. The chain has a weight, and pulling it up involves working against gravity.
To calculate this, we first identify the mass of the hanging part of the chain. Since one-fourth of the chain is hanging, the mass is \(\frac{m}{4}\). This mass experiences the gravitational force downward.
The work required, in general, to lift an object is given by the formula: \[ \text{Work} = \text{Force} \times \text{Distance} \]
For the hanging chain, the force is its weight (mass times gravitational acceleration, \(g\)). The work done against gravity is then the product of this force and the distance through which the mass is lifted.
center of mass
The center of mass is a point representing the mean position of the matter in a body or system. For symmetrical objects and uniform distributions, it's at the geometric center.
In the context of this problem, we consider the hanging part of the chain and its center of mass. Since the hanging part is one-fourth of the chain, the center of mass of this segment is halfway along its length. This is because mass is distributed uniformly along the chain.
So, if the length of the chain is \(L\), the length of the hanging part is \(\frac{L}{4}\), and its center of mass is located at \(\frac{L}{8}\) from the edge of the table. When pulling up, we lift this center of mass through a vertical distance of \(\frac{L}{8}\).
Understanding the center of mass helps us determine the distance over which to calculate the work done against gravity.
physics problem-solving
Physics problem-solving often involves breaking down complex problems into simpler fundamental concepts. In this problem, we follow a straightforward approach to find the work needed to pull the chain up.
Here are the steps broken down:
  • Understand the problem: Identify what is being asked. Here, it鈥檚 about finding the work required to pull one-fourth of the chain onto the table.
  • Determine the mass of the segment: Calculate the mass of the part of the chain being moved, which is \(\frac{m}{4}\).
  • Find the center of mass: Determine the position of the center of mass of the hanging segment to find the vertical distance it travels, which is \(\frac{L}{8}\).
  • Set up the work integral: Combine the force due to weight (gravity) with the distance moved to calculate work.
  • Simplify and solve: Use the formula \[ \text{Work} = \frac{m \times g \times L}{32} \] to find the final expression.
Breaking problems down in this way makes even complex physics more manageable and understandable.

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Most popular questions from this chapter

. Playground Slide A girl whose weight is \(267 \mathrm{~N}\) slides down a \(6.1 \mathrm{~m}\) playground slide that makes an angle of \(20^{\circ}\) with the horizontal. The coefficient of kinetic friction between slide and child is \(0.10\). (a) How much energy is transferred to thermal energy? (b) If the girl starts at the top with a speed of \(0.457 \mathrm{~m} / \mathrm{s}\), what is her speed at the bottom?

Mount Everest The summit of Mount Everest is \(8850 \mathrm{~m}\) above sea level. (a) How much energy would a \(90 \mathrm{~kg}\) climber expend against the gravitational force on him in climbing to the summit from sea level? (b) How many candy bars, at \(1.25 \mathrm{MJ}\) per bar, would supply an energy equivalent to this? Your answer should suggest that work done against the gravitational force is a very small part of the energy expended in climbing a mountain.

Diatomic Molecule The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$ U=\frac{A}{r^{12}}-\frac{B}{r^{6}} $$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation-that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

Spring at the Top of an Incline a spring with spring constant \(k=170 \mathrm{~N} / \mathrm{m}\) is at the top of a \(37.0^{\circ}\) frictionless incline. The lower end of the incline is \(1.00 \mathrm{~m}\) from the end of the spring, which is at its relaxed length. A \(2.00 \mathrm{~kg}\) canister is pushed against the spring until the spring is compressed \(0.200 \mathrm{~m}\) and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

Frictionless Ramp In Fig. \(10-64\), block \(A\) of mass \(m_{A}\) slides from rest along a frictionless ramp from a height of \(2.50 \mathrm{~m}\) and then collides with stationary block \(B\), which has mass \(m_{B}=2.00 m_{A} .\) After the collision, block \(B\) slides into a region where the coefficient of kinetic friction is \(0.500\) and comes to a stop in distance \(d\) within that region. What is the value of distance \(d\) if the collision is (a) elastic and (b) completely inelastic?
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