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Block on inclined plane

  • Thread starter bulldog23
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  • #1
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Homework Statement



A block with mass m = 23.4 kg slides down an inclined plane of slope angle 31.1 o with a constant velocity. It is then projected up the same plane with an initial speed 1.50 m/s. How far up the incline will the block move before coming to rest?



Homework Equations


KE=1/2mv^2
W=F*displacement*cos(theta)


The Attempt at a Solution


I don't really know how to even start this problem. Can someone walk me through it?
 

Answers and Replies

  • #2
learningphysics
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Use the coming downward part to find out the frictional force. There are different ways of doing this...

one ways is: work by friction = change in mechanical energy.

once you know the frictional force, you can do the next part.
 
  • #3
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I haven't ever done anything with mechanical energy. How do you figure that out?
 
  • #4
learningphysics
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I haven't ever done anything with mechanical energy. How do you figure that out?
hmmm... actually I was wrong.... that's probably not the best way to do the first part... probably best to just use forces...

The net force acting on the block as it slides downwards is 0 (since acceleration is constant). Can you use this to figure out the frictional force?
 
  • #5
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so then frictional force must be equal but opposite to mg??
 
  • #6
learningphysics
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so then frictional force must be equal to mg??
not mg... what is the component of mg along the plane?
 
  • #7
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I'm not quite sure. Is it just g?
 
  • #8
learningphysics
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I'm not quite sure. Is it just g?
no... mg divides into two components... one perpendicular to the plane... one parallel to the plane.

use the angle 31 degrees.
 
  • #9
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m*cos(theta)
 
  • #10
learningphysics
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  • #11
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alright, my bad I meant to put mgcos(theta), but I see why it's mgsin(theta). So then I figure that out and divide by 1.5 m/s?
 
  • #12
learningphysics
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alright, my bad I meant to put mgcos(theta), but I see why it's mgsin(theta). So then I figure that out and divide by 1.5 m/s?
well... mgsin(theta) is the frictional force... the magnitude doesn't change when going up or down... but the direction is different...

So when the object is going up, what is the net force acting on it along the plane?

Take the net force... divide by m... that's the acceleration.
 
  • #13
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when going up would it be -mgsin(theta)?
 
  • #14
learningphysics
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when going up would it be -mgsin(theta)?
yes, that's the frictional force... but you've also got -mgsin(theta) due to gravity... so it's a total of -2mgsin(theta).
 
  • #15
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so when I plug the numbers in I get -6.74 m. Is this right?
 
  • #16
learningphysics
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so when I plug the numbers in I get -6.74 m. Is this right?
I don't think so. How did you get that?
 
  • #17
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-2mgsin(theta)/m
 
  • #18
learningphysics
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-2mgsin(theta)/m
so that's -2gsin(theta). I get -10.12m/s^2
 
  • #19
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oh ok. So then I divide by the initial speed to get how far up it travels?
 
Last edited:
  • #20
learningphysics
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oh ok. So then I divide by the initial speed?
No. You need to get the distance... you have v0, a and vf = 0. use kinematics to get the distance.
 
  • #21
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alright, thanks for the help!
 
  • #22
learningphysics
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alright, thanks for the help!
Did you get the distance?
 

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