Incline Plane / Friction Problem

  • Thread starter Destrio
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  • #1
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A block slides down an inclined plane of slope angle A with constant velocity. It is then projected up the same plane with an initial speed Vo.
a) How far up the incline will it move before coming to rest?
b) Will it slide down again?

So when it comes at rest, Vy = 0, Vx = 0
Since it slides down at a constant velocity, the angle must be such that the force of friction allows it to slide down without an applied force. So I believe it will slide down again.

y = Voyt - (1/2)gt^2
x = Voxt

tanA = y/x

I'm not too sure how to proceed from here, I know I will have to solve for time to get the distance in other terms
 

Answers and Replies

  • #2
Dick
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Mmm. The angle is such that the force of friction coming down is EQUAL to the applied force which is mg*sin(A). The forces are equal and opposite. Going up, the frictional force and gravity will be acting in the same direction. Both down. So how high would it go compared with the frictionless case?
 
  • #3
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I believe both cases involve friction, the first just implying that the frictional force must be equal to the applied force (gravity) for it to be sliding at a constant velocity.

So going back up the forces working against it will be friction and gravity, like you said working in the same direction.

Since it is the same plane, is it safe to assume that the force of friction is still equal to the force of gravity?

Fnet = Ff + Fg
Fnet = mg*sin(A) + mg*sin(A)
Fnet = 2mg*sin(A)
 
  • #4
Dick
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Yes, if you agree with what I said.
 
  • #5
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Does that mean what you said could be false? :p

so
Fnet = Ff + Fg
Fnet = mg*sin(A) + mg*sin(A)
Fnet = 2mg*sin(A)

with that i can solve for acceleration

Fnet = ma
ma = 2mg*sin(A)
a = 2g*sin(A)

Vf^2 = Vo^2 + 2ad
0 = Vo^2 + 2(2g*sin(A))d
d = -Vo^2 /4g*sin(A)

Can I do that or do I have to mind the x and y coordinates?

Thanks
 
  • #6
Dick
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What I said could always be false. I do make mistakes. :P And if you use the form you did (energy conservation) then you don't need to worry about x and y. It looks correct to me (if you are taking g to be negative). Note that's 1/2 what you would get if the plane were frictionless. As you have twice the force acting.
 
  • #7
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Excellent
Thanks for your help :)
 

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