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Chapter 11: Problem 18

A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Friction is absent. The spring constant of the spring is 3600 N/m. The piston has a negligible mass and a radius of 0.024 m. (a) When the 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
Compression is 0.0509 m, and work done is 4.66 J.

Step by step solution

01

Understand the Given Information

We are given that the piston in the cylinder has a radius of 0.024 m, the spring constant is 3600 N/m, and there is no friction. Atmospheric pressure outside is acting on the piston.
02

Calculate the Force Due to Atmospheric Pressure

Atmospheric pressure, denoted as \( P_a \), is approximately 101325 Pa. To find the force \( F \) applied by atmospheric pressure, use the formula for pressure: \( F = P_a \times A \), where \( A \) is the area of the piston surface.The area \( A \) of the circular piston is calculated using \( A = \pi r^2 \), with \( r = 0.024 \, \text{m} \). So \( A = \pi \times (0.024)^2 \approx 0.00181 \, \text{m}^2 \). Then, \( F = 101325 \, \text{Pa} \times 0.00181 \, \text{m}^2 \approx 183.4 \, \text{N} \).
03

Use Hooke's Law to Find Compression

Hooke's Law relates the force made by a spring to the displacement (compression) \( x \) when a force \( F \) is applied: \( F = kx \), where \( k \) is the spring constant.Rearrange to find \( x \): \( x = \frac{F}{k} \). Substitute the values: \( x = \frac{183.4 \, \text{N}}{3600 \, \text{N/m}} \approx 0.0509 \, \text{m} \).
04

Calculate the Work Done by Atmospheric Pressure

Work done by a force is given by the formula: \( W = \frac{1}{2}kx^2 \), where \( x \) is the spring compression we've calculated.Substitute in the values: \( W = \frac{1}{2} \times 3600 \, \text{N/m} \times (0.0509 \, \text{m})^2 \approx 4.66 \, \text{J} \).

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Key Concepts

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

Work and Energy
Work and energy are fundamental concepts in physics that help us understand forces and motion. When work is done on an object, energy is transferred to or from that object. This is expressed in the formula for work: \[ W = F imes d \] where \( W \) is the work done, \( F \) is the force applied, and \( d \) is the distance over which the force is applied. In our example, the work done by atmospheric pressure on the piston compresses the spring, storing potential energy in it.
  • Energy can be either kinetic (due to motion) or potential (stored energy).
  • Since the spring doesn't move, but stores energy, the work done appears as potential energy.
Understanding work and energy in this context helps explain why the spring compresses when pressure is applied, and how the energy changes form.
Spring Constant
The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It tells us how much force is needed to stretch or compress a spring by a unit of length. It's expressed in units of \( \, \text{N/m} \). In our exercise, the spring constant is 3600 N/m. This means a force of 3600 N is required to compress the spring by 1 meter.
  • A larger spring constant indicates a stiffer spring.
  • A smaller spring constant means the spring is less stiff and easier to compress.
Knowing the spring constant is essential to determine how much a spring will compress under a given force, using Hooke's Law. This helps us solve for the displacement \( x \) caused by atmospheric pressure in our exercise.
Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of air in the atmosphere. It is typically measured in Pascals (Pa). Standard atmospheric pressure at sea level is approximately 101325 Pa. In our problem, this constant pressure acts on the piston, leading to the compression of the spring inside the cylinder.
  • Atmospheric pressure can vary with altitude and weather conditions.
  • It acts on surfaces exposed to the air, like the piston in the cylinder.
By calculating the force exerted by the atmospheric pressure on the piston's surface, we can understand how it leads to the compression of the spring, effectively demonstrating energy transfer through work.
Hooke's Law
Hooke's Law is a principle of physics that states that the force needed to extend or compress a spring by a distance \( x \) is proportional to that distance. It's expressed as:\[ F = kx \]where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the spring's equilibrium position.
  • Hooke's Law applies only in the elastic region, where the spring returns to its original shape after the force is removed.
  • If stretched too far, the spring may not return, and Hooke's Law no longer applies.
In our solution, Hooke's Law was used to calculate the compression \( x \) of the spring caused by the atmospheric force. This allowed us to determine how much the spring compresses under a given force, providing insight into the system's behavior.

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Most popular questions from this chapter

A hydrometer is a device used to measure the density of a liquid. It is a cylindrical tube weighted at one end, so that it floats with the heavier end downward. The tube is contained inside a large 鈥渕edicine dropper,鈥� into which the liquid is drawn using the squeeze bulb (see the drawing). For use with your car, marks are put on the tube so that the level at which it floats indicates whether the liquid is battery acid (more dense) or antifreeze (less dense). The hydrometer has a weight of \(W=5.88 \times 10^{-2} \mathrm{N}\) and a cross-sectional area of tw \(A=7.85 \times 10^{-5} \mathrm{m}^{2} .\) How far from the bottom of the tube should the mark be put that denotes (a) battery acid \(\left(\rho=1280 \mathrm{kg} / \mathrm{m}^{3}\right)\) and (b) antifreeze \(\left(\rho=1073 \mathrm{kg} / \mathrm{m}^{3}\right) ?\)

A \(58-\mathrm{kg}\) skier is going down a slope oriented \(35^{\circ}\) above the horizontal. The area of each ski in contact with the snow is 0.13 \(\mathrm{m}^{2}\) Determine the pressure that each ski exerts on the snow.

A siphon tube is useful for removing liquid from a tank. The siphon tube is first filled with liquid, and then one end is inserted into the tank. Liquid then drains out the other end, as the drawing illustrates. (a) Using reasoning similar to that employed in obtaining Torricelli鈥檚 theorem (see Example 16), derive an expression for the speed \(v\) of the fluid emerging from the tube. This expression should give \(v\) in terms of the vertical height \(y\) and the acceleration due to gravity \(g\) . (Note that this speed does not depend on the depth \(d\) of the tube below the surface of the liquid.) (b) At what value of the vertical distance y will the siphon stop working? (c) Derive an expression for the absolute pressure at the highest point in the siphon (point \(A )\) in terms of the atmospheric pressure \(P_{0},\) the fluid density \(\rho, g,\) and the heights \(h\) and \(y\) (Note that the fluid speed at point \(A\) is the same as the speed of the fluid emerging from the tube, because the cross-sectional area of the tube is the same everywhere.)

A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is \(1.80 \times 10^{5} \mathrm{Pa}\) . Assuming that the top and bottom surfaces of the cap each have an arew of \(4.10 \times 10^{-4} \mathrm{m}^{2}\) , obtain the magnitude of the force that the screw thread exerts on the cap in order to keep it on the bottle. The air pressure outside the bottle is one atmosphere.

Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is closed. Both containers are filled initially to the same height of 1.00 m, one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.
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