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Chapter 11: Problem 88

A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open at the top. Friction is absent. The spring constant of the spring is \(3600 \mathrm{~N} /\) \(\mathrm{m}\). The piston has a negligible mass and a radius of \(0.025 \mathrm{~m}\). (a) When air beneath the piston is completely pumped out, how much does the atmospheric pressure cause the spring to compress? (b) How much work does the atmospheric pressure do in compressing the spring?

Short Answer

Expert verified
The spring compresses by 0.0552 m and 5.48 Joules of work is done by the atmospheric pressure.

Step by step solution

01

Understanding the problem

When the air is pumped out from beneath the piston, the atmospheric pressure outside will push the piston downwards, compressing the spring. The atmospheric pressure is given as the standard atmospheric pressure, which is approximately 101,325 Pascals (Pa). We need to determine the compression of the spring at this pressure.
02

Calculating the force due to atmospheric pressure

First, calculate the force exerted by atmospheric pressure on the piston using the formula:\[ F = P \times A \]where \( P \) is atmospheric pressure (101,325 Pa) and \( A \) is the area of the piston:\[ A = \pi r^2 = \pi (0.025)^2 \]
03

Solving for Force using the given data

Substitute the values to find the area:\[ A = \pi (0.025)^2 \approx 0.00196 \text{ m}^2 \]Now, calculate the force:\[ F = 101,325 \times 0.00196 \approx 198.60 \text{ N} \]
04

Finding the spring compression using Hooke's Law

According to Hooke's Law, the force exerted by the spring is:\[ F = k \times x \]where \( k \) is the spring constant (3600 N/m) and \( x \) is the compression of the spring.Set the atmospheric force equal to the spring force and solve for \( x \):\[ 198.60 = 3600 \times x \]\[ x = \frac{198.60}{3600} \approx 0.0552 \text{ m} \]
05

Calculating the work done by atmospheric pressure

The work done by atmospheric pressure in compressing the spring can be calculated by:\[ W = \frac{1}{2} k x^2 \]where \( x = 0.0552 \) m. Substituting the values gives:\[ W = \frac{1}{2} \times 3600 \times (0.0552)^2 \approx 5.48 \text{ Joules} \]

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Key Concepts

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

Atmospheric pressure
Atmospheric pressure is the force exerted by the weight of the atmosphere above us. This pressure is vital in various physical phenomena and calculations. At sea level, the standard atmospheric pressure is around 101,325 Pascals (Pa), also known as 1 atmosphere (atm). Atmospheric pressure affects everyday experiences, such as feeling pressure on your ears during a flight.

In the piston-spring system, atmospheric pressure plays a key role when the air beneath the piston is removed. This pressure forces the piston downwards, compressing the spring. Understanding atmospheric pressure is crucial for solving physics problems involving force and motion. It helps in determining how much force is exerted on the piston and thus, how the spring behaves under these conditions.
Hooke's Law
Hooke's Law is a fundamental principle in physics that explains how springs and elastic materials behave. It states that the force required to compress or stretch a spring is directly proportional to the distance it is compressed or stretched. This relationship is mathematically described by the formula:
  • \( F = k \times x \)
where \( F \) is the force applied, \( k \) is the spring constant (a measure of the spring's stiffness), and \( x \) is the displacement from the spring's equilibrium position.

In the context of the piston-spring system, Hooke鈥檚 Law allows us to relate the force exerted by the atmospheric pressure to the compression of the spring. By knowing the spring constant (3600 N/m), we can calculate the extent to which the spring is compressed when subjected to a specific force, such as atmospheric pressure.
Piston-spring system
A piston-spring system is a mechanical setup where a piston compresses a spring when subjected to an external force. This system is commonly used in engines, hydraulics, and shock absorbers. In this exercise, the system involves a cylinder fitted with a piston atop a spring. The piston can move freely without any friction, allowing for seamless compression and decompression of the spring.

The absence of air beneath the piston creates a vacuum, allowing atmospheric pressure to push the piston down and compress the spring. With a given spring constant, we can analyze how much the spring will compress under these conditions. Understanding the dynamics of the piston-spring system is essential for predicting mechanical behavior and designing efficient mechanical components.
Work done calculation
Work is a measure of energy transfer when a force is applied over a distance. In physics, the work done by a force is calculated using the formula:
  • \( W = F \times d \cos(\theta) \)
For springs and when dealing with Hooke's Law, the work done by a spring force is determined using:
  • \( W = \frac{1}{2} k x^2 \)
where \( k \) is the spring constant and \( x \) is the compression length of the spring.

In this piston-spring scenario, we are calculating the work done by atmospheric pressure in compressing the spring. The formula reflects the energy stored in the spring as it is compressed, based on the distance \( x \) it is compressed and the inherent stiffness of the spring \( k \). This concept is vital in understanding energy conversion and storage in mechanical systems.

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Most popular questions from this chapter

The human lungs can function satisfactorily up to a limit where the pressure difference between the outside and inside of the lungs is one-twentieth of an atmosphere. If a diver uses a snorkel for breathing, how far below the water can she swim? Assume the diver is in salt water whose density is \(1025 \mathrm{~kg} / \mathrm{m}^{3}\)

The main water line enters a house on the first floor. The line has a gauge pressure of \(1.90 \times 10^{5} \mathrm{~Pa}\). (a) A faucet on the second floor, \(6.50 \mathrm{~m}\) above the first floor, is turned off. What is the gauge pressure at this faucet? (b) How high could a faucet be before no water would flow from it, even if the faucet were open?

One of the concrete pillars that support a house is \(2.2 \mathrm{~m}\) tall and has a radius of \(0.50 \mathrm{~m}\). The density of concrete is about \(2.2 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). Find the weight of this pillar in pounds \((1 \mathrm{~N}=0.2248 \mathrm{lb})\).

A suitcase (mass \(\mathrm{m}=16 \mathrm{~kg}\) ) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures \(0.50 \mathrm{~m}\) by \(0.15 \mathrm{~m}\). The elevator is moving upward, the magnitude of its acceleration being \(1.5 \mathrm{~m} / \mathrm{s}^{2}\). What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?

At reviews the concept that plays the central role in this problem. (a) The volume flow rate in an artery supplying the brain is \(3.6 \times 10^{-6} \mathrm{~m}^{3} / \mathrm{s}\). If the radius of the artery is \(5.2 \mathrm{~mm}\), determine the average blood speed. (b) Find the average blood speed at a constriction in the artery if the constriction reduces the radius by a factor of \(3 .\) Assume that the volume flow rate is the same as that in part (a).
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