Force and acceleration

  • Thread starter Vanity
  • Start date
  • #1
14
0
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 wich 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 physic english terms.. :uhh:
 
Last edited:

Answers and Replies

  • #2
1,426
1
What is the lenght of one mark?
 
Last edited:
  • #3
14
0
Werg22 said:
What is the lenght of one mark?
the lenght 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
 
  • #4
1,426
1
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
 
Last edited:
  • #5
14
0
Werg22 said:
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 alot ;)
 
  • #6
1,426
1
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).
 
Last edited:

Related Threads on Force and acceleration

  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
922
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
13
Views
2K
Top