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

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Homework Help Overview

The discussion revolves around a physics problem involving a wood block launched up an inclined ramp, specifically focusing on Newton's Laws, friction, and the calculations related to height and speed on an incline.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of free body diagrams and the resolution of forces into components. There are attempts to derive equations for net forces and acceleration, as well as questions about how these relate to height and speed.

Discussion Status

Some participants are exploring the relationships between forces acting on the block, while others are questioning the accuracy of their calculations and the application of Newton's Laws. There is an ongoing dialogue about the correct approach to find height and acceleration, with some guidance provided regarding the simplification of equations.

Contextual Notes

Participants mention potential misunderstandings in the application of forces and the need for clarity in the equations used. There is also a reference to external resources for further clarification.

sktgurl930
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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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