Atmospheric pressure and the spring constant

In summary, the problem discusses a cylinder with a piston and spring, and asks about the compression of the spring due to the atmospheric pressure when the air is completely pumped out. The answer for part (a) is 0.0214 m. For part (b), using the work equation, the result is 1.969 J, but there may be a mistake in the problem as the work done on the spring may only be half of this due to the assumption that the piston has zero mass.
  • #1
rasputin66
12
0

Homework Statement


I already found A, but I keep getting wrong answers for B.

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 4300 N/m. The piston has a negligible mass and a radius of 0.017 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?

k= 4300 N/m
r= 0.017 m
A= 0.00091 m^2
P= 101 kN/m
F= 91.995 N

Homework Equations


P= F/A
W= F d

The Attempt at a Solution



I found the answer to A which is 0.0214 m. So then I tried to use the Work equation and got 1.969 J. But this is wrong. I'm stuck! Please help!
 
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  • #2
Please post your working.
 
  • #3
OK, well the spring moved 0.0214 m and the F involved is 91.995 N.
W = F d
W = (91.995 N) (0.0214 m)= 1.968693 J
The homework program on the computer is telling me this is wrong. I don't know why. Do you think this is OK? Sometimes the program is mistaken.
 
  • #4
That all seems correct, though my numbers are slightly different: .02133m and 1.956J.
But I wonder if the problem setter has made a mistake here. If you calculate the work done on the spring, kx2/2, you will only get half as much.
Imagine all the air being drawn out very quickly. At first, the force from the atmospheric pressure will be much more than the compressive force in the spring. Since all masses are taken to be zero, this will result in infinite acceleration! When that sort of thing happens to equations you have to relax your assumptions to reintroduce some sanity. In this case, allow the piston to have some small mass.
Now the piston will acquire KE and momentum. At the point where the spring has reached the equilibrium compression you calculated, 0.0213m, half of the work done by the atmosphere is in the form of KE of the piston. Note that this is true no matter how light we make the piston. In a real environment, the spring will then oscillate, losing energy to friction, asymptotically converging to the 0.0213 displacement.
In short, the work the atmosphere does will be about 2J, but the work then stored in the spring will be only about 1J.
 
  • #5


I would first check my calculations and make sure I am using the correct units. In this case, the unit for pressure should be in pascals (Pa) and the unit for work should be in joules (J). It looks like you may have accidentally used kilonewtons (kN) instead of newtons (N) in your calculations.

Next, I would consider the relationship between atmospheric pressure and the spring constant. Atmospheric pressure is the force exerted by the weight of the air above a certain area. In this case, the area is the cross-sectional area of the piston, which is 0.00091 m^2. The force of atmospheric pressure can be calculated by multiplying the pressure (in Pa) by the area (in m^2).

Since the spring is being compressed by the atmospheric pressure, the work done by the atmospheric pressure is equal to the change in potential energy of the spring. This can be calculated using the equation W = ½kx^2, where k is the spring constant and x is the distance the spring is compressed.

Therefore, for part (b), we can calculate the work done by the atmospheric pressure as W = ½(4300 N/m)(0.0214 m)^2 = 2.03 J.

If you are still getting a different answer, I would recommend double-checking your calculations and units. If you are still stuck, you can always consult with a classmate or your instructor for further clarification.
 

1. What is atmospheric pressure and how is it measured?

Atmospheric pressure refers to the force exerted by the weight of the Earth's atmosphere on a given area. It is typically measured using a barometer, which calculates the pressure based on the height of a column of mercury or aneroid chamber.

2. How does atmospheric pressure affect weather patterns?

High atmospheric pressure usually indicates calm and clear weather, while low atmospheric pressure can lead to stormy weather. Changes in atmospheric pressure can also affect wind patterns and precipitation.

3. What is the spring constant and why is it important?

The spring constant, also known as the force constant, is a measure of how easily a spring can be compressed or stretched. It is important in the study of atmospheric pressure because it helps to determine the strength and elasticity of materials that are affected by changes in pressure.

4. How does the spring constant affect the behavior of gases in the atmosphere?

The spring constant plays a role in the ideal gas law, which describes the behavior of gases in relation to pressure, volume, and temperature. This law states that as pressure increases, volume decreases, and vice versa, with the spring constant being a factor in this relationship.

5. Can atmospheric pressure be manipulated or controlled?

Atmospheric pressure is a natural phenomenon that cannot be controlled or manipulated on a large scale. However, it can be affected by human activity on a smaller scale, such as in the controlled environment of a laboratory or in the construction of buildings and structures.

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