Problem 88
Question
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?
Step-by-Step Solution
Verified Answer
The spring compresses by 0.0552 m and 5.48 Joules of work is done by the atmospheric pressure.
1Step 1: 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.
2Step 2: 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:\[ A = \pi r^2 = \pi (0.025)^2 \]
3Step 3: 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:\[ F = 101,325 \times 0.00196 \approx 198.60 \text{ N} \]
4Step 4: 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 \( x \):\[ 198.60 = 3600 \times x \]\[ x = \frac{198.60}{3600} \approx 0.0552 \text{ m} \]
5Step 5: 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:\[ W = \frac{1}{2} \times 3600 \times (0.0552)^2 \approx 5.48 \text{ Joules} \]
Key Concepts
Atmospheric pressureHooke's LawPiston-spring systemWork done calculation
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.
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:
In the context of the piston-spring system, Hooke’s 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.
- \( F = k \times x \)
In the context of the piston-spring system, Hooke’s 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.
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:
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.
- \( W = F \times d \cos(\theta) \)
- \( W = \frac{1}{2} k x^2 \)
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.
Other exercises in this chapter
Problem 86
(a) The mass and radius of the sun are \(1.99 \times 10^{30} \mathrm{~kg}\) and \(6.96 \times 10^{8} \mathrm{~m}\). What is its density? (b) If a solid object i
View solution Problem 87
Concept Simulation 11.1 at reviews the central idea in this problem. In an adjustable nozzle for a garden hose, a cylindrical plug is aligned along the axis of
View solution Problem 91
Mercury is poured into a tall glass. Ethyl alcohol is then poured on top of the mercury until the height of the ethyl alcohol itself is \(110 \mathrm{~cm}\). Th
View solution Problem 92
An irregularly shaped chunk of concrete has a hollow spherical cavity inside. The mass of the chunk is \(33 \mathrm{~kg}\), and the volume enclosed by the outsi
View solution