Calculating Speed and Time for a Sliding Disk on an Elevated Air Table

[Vanity]

Hi all!

I'd like to know fi anybody can help me out with this, I just need some info so I can do it by myself.. formulas or something.

I have a disk (looks like a hockey puck) sliding on an air table which is elevated with an angle of 5°. [weight of the disk: 0,558g] Every 0,1s, the disk makes a mark on a sheet of paper.

I need to make a table of the position (y) and time (x) and another one with average speed and time spent between each points (wich is 0,1s) .. anyone know how to do that?

Thanks for the help ;)
- Alex

ps: Sorry, I'm not too use with physics english terms..

 

[Werg22]

What is the length of one mark?

 

[Vanity]

What is the length of one mark?

the length is smaller between each points, here's what it looks like:

Between Points...
[1-2] 7,0 cm
[2-3] 6,2 cm
[3-4] 5,9 cm
[4-5] 5,0 cm
[5-6] 3,9 cm
[6-7] 3,3 cm
[7-8] 2,3 cm

There we go ;) sorry I didn't mention this in the first post

 

[Werg22]

So between point 1 and 2, v1=0 and v2=2d+v1/t. Vavg: v2-v1/2.

So I'll only do 1-2:

v1=0
v2=2(0.07)+0/0.1=1.4 m/s
vavg=1.4-0/2=0.7 m/s

 

[Vanity]

So between point 1 and 2, v1=0 and v2=2d+v1/t. Vavg: v2-v1/2.

So I'll only do 1-2:

v1=0
v2=2(0.07)+0/0.1=1.4 m/s
vavg=1.4-0/2=0.7 m/s

Allright, I got it. Thanks a lot ;)

 

[Werg22]

The problem is not true when checking:

vf^2=2(0.07)sin5*g=0.35 m/s
so vf=1.15 m/s

and vavg=0.35+0/2=0.175 m/s

The time isn't 0.1 s unless the coefficient of friction is 16.37 (wich no matter has).