# Air Pressure and Spring Question

## Homework Statement

What work is done by the air on the spring?

k(The force constant)=640 N/m
x=2.0cm=0.020m
The air is coming out of the tire, into a pressure gauge pushing a spring 2.0cm.

Fx=x*k
W=F*D

## The Attempt at a Solution

First of all, I need to find the force in the equation of work.

Fx= (0.020m)(6.4x2N/m)
Force needs to be positive so I don't use the negative version of the equation since work done by the air needs to be positive thus needing a positive force.

Fx=12.8N

W=(12.8N)(0.020m)=0.26J

The work done by the air on the spring is approx 0.26J

Could someone verify my solution?

Related Introductory Physics Homework Help News on Phys.org
gneill
Mentor
Is the force constant as the spring is pushed?

Is the force constant as the spring is pushed?
I believe so. The book has not told me otherwise.

gneill
Mentor
So, as the spring is being compressed (or stretched), the force it produces is constant?

So, as the spring is being compressed (or stretched), the force it produces is constant?
The spring is being compressed (Or Pushed) by the air, creating a reading on the gauge.

gneill
Mentor
The spring is being compressed (Or Pushed) by the air, creating a reading on the gauge.
No doubt it is. But that doesn't address the question of how much force the spring provides as it is compressed or stretched.

Ask yourself what the spring constant "means" physically, and then think about it in terms of Newton's 3rd law.

No doubt it is. But that doesn't address the question of how much force the spring provides as it is compressed or stretched.

Ask yourself what the spring constant "means" physically, and then think about it in terms of Newton's 3rd law.
The spring constant means.

640 N/m : 640 Newtons required to compress 1 meter.

If its being compressed 0.02m, then the force needed to compress is 12.8N?

Hm.. the resulting force from the spring will be equal to the opposite, -12.8N since the air is building some Potential energy in the spring.

I can't think of where I may of gone wrong.

gneill
Mentor
The spring constant means.

640 N/m : 640 Newtons required to compress 1 meter.
That's 640 Newtons at the end of the compression of 1 meter, to hold the spring in place. The spring pushes back with that force.

Newtons aren't like buckets of water; They don't accumulate or hang around over time, unlike energy, which can be stored.

How much force is required at 1/2 meter? 1/4 meter? All along the compression path, the force required is different.

Now, it may look as though this is going to complicate your like greatly, but there's another property of the spring that you can take advantage of, namely the energy stored in it.

Since the airflow/pressure is constant as the tire gauge is inserted into the tire then the spring is held in place.

gneill
Mentor
Since the airflow/pressure is constant as the tire gauge is inserted into the tire then the spring is held in place.
No doubt that's also true. But to get to that place, there was a history of movement of the spring. At each point along that path the spring's reaction force was different.

Suppose you had to push a box along level ground for 10 meters, requiring a force of 10N to keep it moving at a constant velocity, and then up a slope for another 2 meters, requiring 150N of force to keep it moving. At the end of your journey you're holding the box in place by leaning against it with a continual force of 150N. How much work was done to move the box the total of 12 meters? It's not 150N x 12m. The 150N was only applied for a small portion of the journey; the first portion was easy: only 10N applied the whole way.

So you see, the force at the end of the journey does not tell you what the force was along the whole journey. That's why work done is the integral of f(x)dx .

There is something in the system under consideration that does 'remember' the history of the work done though. Can you think of what that might be?