Calculating Coefficient of Friction with Practical Investigation Results

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To calculate the coefficient of friction from the practical investigation involving a rubber band and a puck, first ensure the rubber band is calibrated to obtain a reliable spring constant (k). The elastic potential energy stored in the band can be equated to the kinetic energy of the puck, allowing for the calculation of acceleration. The friction force can then be expressed in terms of the normal force and the coefficient of friction, leading to the equation μmg = ma, where mass cancels out. It's important to note that the rubber band's behavior can vary significantly with extension, so maintaining a consistent range of extension is crucial for accurate results. Using an inclined plane can also provide a method to determine static and dynamic friction coefficients for verification.
Adem
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So I've done a practical investigation.
Pretty much I used a elastic band to launch a puck on a horizontal floor. I have measured the extension and force for the elastic band, and the extension and stopping distance for the puck on two different surfaces.
How do I calculate the coefficient of friction?
thanks.
 
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Adem said:
How do I calculate the coefficient of friction?
Actually, there may be two of those: one as long as it's not moving and one if it is (the latter is usually smaller).
Did you calibrate your rubber band ? Is it a bit linear (like an ideal spring: extension = k ##\times## force) ?

Are you familiar with friction coefficient formulas like ##F_{\rm friction} = \mu F_{\rm normal} = \mu\, mg ## ?
 
Yeah, it's linear.
 
I have a force vs ext for the band, but don't have the force applied when I used it for the puck being launched.
 
You must calibrate the band for obtain k.

Then you need to suposse that all elastic energy of the band is transferred to the puck in the form of kinetic energy. Then: $$\frac 1 2 k·x^2= \frac 1 2 m·v_0^2$$

Then you can calculate the aceleration $$0=v_0^2+2·a·distance$$

During the mouvement there are only friction force, then: $$\mu·m·g=m·a$$
(mass cancel)
I don't know if there are some wrong in this procedure, please say me if this procedure is not correct. Sorry for my English.
 
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Even in a moore compact way: Eband=Wfriction $$\frac 1 2 k·x^2=\mu·m·g·distance$$
 
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Thanks guys!
 
Static friction can be easily found with an inclined plane. (you can look this up) With a bit of care, you can also use an inclined plane to find the limiting slope that will keep the puck moving when nudged, which will give you the dynamic friction. (Good as a cross check if nothing else.)

The k of a rubber band can vary a lot over the full range of extensions. You tend to get a very S shaped curve so you need to operate over a range of expansions that doesn't involve the section where the band goes 'stiff' if you want an accurate value of the Kinetic Energy that it gives the puck. With a bit of care, you can also make sure that the band is not operating with low extension. (A restraining loop of thread, for instance).
 
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