How do Newton's Laws relate to friction and inclined ramps?

AI Thread Summary
The discussion focuses on a physics problem involving a 2.8 kg wood block launched up a 34° inclined ramp with an initial speed of 8.56 m/s. Participants analyze the forces acting on the block, including friction and gravitational components, to determine the vertical height the block reaches and its speed when sliding back down. There is confusion regarding the application of Newton's laws, particularly in resolving forces into components and understanding acceleration in the vertical direction. Suggestions include drawing a free body diagram and clarifying the equations used to solve for height and acceleration. The conversation emphasizes the importance of correctly applying physics principles to arrive at the correct solution.
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Homework Statement



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?


Homework Equations



Fk=mk*N
a=Fk/M

The Attempt at a Solution


Fk=.20*2.8*9.8
=5.488

a=1.96
and yea i really don't kno wat to do with this
 
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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.
 
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
 
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?
 
sktgurl930 said:
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

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.
 
sktgurl930 said:
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?

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?
Xf=xi+(Vf^2-Vi^2/2a)= 9.133
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
 
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