# Homework Help: Newtons laws friction

1. Sep 23, 2008

### sktgurl930

1. The problem statement, all variables and given/known data

A 2.8 kg wood block is launched up a wooden ramp that is inclined at a 34° angle. The block’s initial speed is 8.56 m/s. What vertical height does the block reach above its starting point?

What speed does it have when it slides back down to its starting point?

2. Relevant equations

Fk=mk*N
a=Fk/M

3. The attempt at a solution
Fk=.20*2.8*9.8
=5.488

a=1.96
and yea i really dont kno wat to do with this

2. Sep 23, 2008

### hage567

Did you draw a free body diagram? You are missing some things. Also, there is an angle involved, so you must resolve your forces into their components.

3. Sep 24, 2008

### sktgurl930

ok so i think i drew it right and this is the equation i got from it
fnetx= -Fk-Wsin+v=Ma
fnety=N-Wcos=Ma

how does that give me height and accerelation

4. Sep 24, 2008

### sktgurl930

ok so this is wat i got ignore the last post
wx=M*gsin = 8.324N
Fk= Mk*N = 3.9396 N
Fk-Wx/m= - 4.385 m/s^2
Xf=xi+(Vf^2-Vi^2/2a)= 9.133
Xf*sin= 3.8599
would this be my height????

5. Sep 24, 2008

### hage567

These are sort of correct, there's a few problems though:

v is not a force, so you can't add it as one in your equation.

If you look in the y direction, there is no movement in it; the block stays on the incline. So the acceleration is zero in that direction. This makes your equation simpler.

Well it will let you solve for acceleration, after which you can work on finding the height.

6. Sep 24, 2008

### hage567

I'm having trouble understanding what you've done here. I don't see how you got 8.324 N. What did you use for N when you found Fk?
This doesn't make sense the way you've typed it.

Take a look at this website, maybe it will help you http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html#nt2cn