# Science fair - Formula for friction to thermal energy?

## Homework Statement

For science fair, I'm doing a project in which we are trying to calculate the force of a thermometer being moved normally, versus a thermometer being slowed down by a high-friction material. The thermometer will (Potentially) read the amount of thermal energy produced by the friction, and we can then calculate the formula between the force of friction, and the thermal energy produced.

## Homework Equations

Has the formula for the relation between the amount of friction and thermal energy already been experimented and noted? (This is my actual question for the PhysicsForums community)

## The Attempt at a Solution

Experimentation (Details in part one).

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berkeman
Mentor

## Homework Statement

For science fair, I'm doing a project in which we are trying to calculate the force of a thermometer being moved normally, versus a thermometer being slowed down by a high-friction material. The thermometer will (Potentially) read the amount of thermal energy produced by the friction, and we can then calculate the formula between the force of friction, and the thermal energy produced.

## Homework Equations

Has the formula for the relation between the amount of friction and thermal energy already been experimented and noted? (This is my actual question for the PhysicsForums community)

## The Attempt at a Solution

Experimentation (Details in part one).
Welcome to the PF.

What is meant by a "thermometer being moved normally"? Thermometers are normally fixed in position.

And as for when you are doing work on the thermometer by pushing it against friction, do you know the equation for Work as a function of Force and Distance? If you know the work done on the thermometer, what can you say about its energy? And how does that relate to its temperature?

By "The thermometer being moved normally", I mean measuring pulling it with a spring scale, without applying friction. Due to the energy having to have an outlet, as the energy does not simply disperse, I'm calculating the transfer from the energy produced by friction, to thermal energy. I do understand the concept of W=F×D, but the expirement will solely be measuring force in Newtons.

berkeman
Mentor
By "The thermometer being moved normally", I mean measuring pulling it with a spring scale, without applying friction. Due to the energy having to have an outlet, as the energy does not simply disperse, I'm calculating the transfer from the energy produced by friction, to thermal energy. I do understand the concept of W=F×D, but the expirement will solely be measuring force in Newtons.
So when you pull it with some force in Newtons against friction for some distance, you do some work which goes into warming both the thermometer and the material you pulled it across. How can you estimage how that warming divides between the thermometer and the friction material? Once you estimate the fraction of the energy that went into warming the thermometer, what other factors do you need to take into account to figure out how much the temperature of the thermometer will increase?

So when you pull it with some force in Newtons against friction for some distance, you do some work which goes into warming both the thermometer and the material you pulled it across. How can you estimage how that warming divides between the thermometer and the friction material? Once you estimate the fraction of the energy that went into warming the thermometer, what other factors do you need to take into account to figure out how much the temperature of the thermometer will increase?
Well, I think my description may have been unclear, but I'm measuring the force of friction by testing 'With Friction' versus 'No Friction', to first, determine the force of the friction by the addition onto the spring scale. When the bottom of the thermometer rubs against the high-friction material, it also creates thermal energy, that will also be recorded. I do this with different forces of friction, and measure the thermal energy and the friction. I then find similarities between the force of friction, and thermal energy.
This makes a formula like: (Force of friction)x N =(Thermal energy)°
Example: 5N×1.345=6.725°, because I found that the similarity was F×1.345

berkeman
Mentor
Well, I think my description may have been unclear, but I'm measuring the force of friction by testing 'With Friction' versus 'No Friction', to first, determine the force of the friction by the addition onto the spring scale. When the bottom of the thermometer rubs against the high-friction material, it also creates thermal energy, that will also be recorded. I do this with different forces of friction, and measure the thermal energy and the friction. I then find similarities between the force of friction, and thermal energy.
This makes a formula like: (Force of friction)x N =(Thermal energy)°
Example: 5N×1.345=6.725°, because I found that the similarity was F×1.345
What's N? You use it in two different ways, I think.

Can you also post the full lab assignment so we can see what-all is involved, and what is expected in the report?

CWatters
Homework Helper
Gold Member
I think it might work...

With some qualification, it might be reasonable to assume that the temperature (T) indicated on the thermometer is proportional to the amount of frictional energy (Ef) liberated.

So you could try writing an equation such as..

T = K*Ef + C

where K and C are some constants that account for unknowns such as heat energy lost to the air or the material the thermometer is being rubbed against. K would also account for the heat capacity of the thermometer. K and C would have appropriate and different units. C might be negative.

In which case you can you repeat the experiment several times and record T and Ef, where Ef = force * distance.

Then plot a graph of T vs Ef and if it's a straight line you can use it to work out the constants K and C.

You would need to take precautions to ensure the thermometer, and indeed the whole rig, had the same stable temperature before each run.

Edit: If it's not a straight line you will have to find a new equation that might explain the relationship.