Physics question?

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animeshadows wrote on 2011-12-14 06:26

[Image: http://puu.sh/aBXj]
Phunkie wrote on 2011-12-14 07:05

I used this website to help me with my homework throughout my last years of high school and all of college.

Try and see if it helps.

http://hyperphysics.phy-astr.gsu.edu/hbase/crstp.html#c1

One thing that should help you is to work out the units in terms of what they represent? I'm not sure.

Let's see...

s^2 / (g * cm)

Seconds is time.
Grams is mass.
Centimeters is distance.

You're looking at:

time^2 / (mass * distance)

time / distance is the inverse of velocity = 1/v, so you have:

1/mass * 1/velocity * time

time/velocity is the inverse of acceleration = 1/a, so you now can simplify and you get:

1/mass * 1/acceleration, which is the inverse of force (F = m*a).

In my opinion, the slope shows you the amount of force (friction, to be specific) needed to stop a certain mass by a certain distance.

Hope that works!
Chockeh wrote on 2011-12-14 09:21

Quote from Phunkie;691860:
I used this website to help me with my homework throughout my last years of high school and all of college.

Try and see if it helps.

http://hyperphysics.phy-astr.gsu.edu/hbase/crstp.html#c1

One thing that should help you is to work out the units in terms of what they represent? I'm not sure.

Let's see...

s^2 / (g * cm)

Seconds is time.
Grams is mass.
Centimeters is distance.

You're looking at:

time^2 / (mass * distance)

time / distance is the inverse of velocity = 1/v, so you have:

1/mass * 1/velocity * time

time/velocity is the inverse of acceleration = 1/a, so you now can simplify and you get:

1/mass * 1/acceleration, which is the inverse of force (F = m*a).

In my opinion, the slope shows you the amount of force (friction, to be specific) needed to stop a certain mass by a certain distance.

Hope that works!

O_O
-Bookmarks-
animeshadows wrote on 2011-12-14 13:49

d = (.5)Mv^2/F

d=KE/F

y = MX+b

(Substitute values in.)

d = M(KE) + b (but D also equals KE/F, therefore...)

d = (1/f) (KE) + 0
-------------------

Thank you for your help Phunkie!
Phunkie wrote on 2011-12-14 20:35

Quote from animeshadows;692027:
d = (.5)Mv^2/F

d=KE/F

y = MX+b

(Substitute values in.)

d = M(KE) + b (but D also equals KE/F, therefore...)

d = (1/f) (KE) + 0
-------------------

Thank you for your help Phunkie!

Oh, totally! I forgot that work (energy) equals force times distance there and decided to solve it the longer way. haha

Your welcome. :)
Akemii wrote on 2011-12-14 20:45

wat did u guyz jst type.