How far does the box travel along the incline before coming to rest?

  • Thread starter prettynerd
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  • #1
prettynerd
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kayyy these are some hwk questions i don't understand. please help :cry:

1. A person has a choice of either pushing or pulling a sled at a constant velocity. Friction is present. If the angle beta is the same in both cases, does it require less force to push or pull? explain.

I believe it'd be equal because it's the same angle. Am I wrong?

2. A cup of coffee is sitting on a table in an airplane that is flying at a constant altitude and a constant velocity. The coefficient of static friction between the cup and the table is 0.30. Suddenly, the plane accelerates, its altitude remaining the same. What is the max acceleration that the plane can have without the cup sliding backward on the table?

3. A box is sliding up an incline plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is 0.180. The initial speed of the box at the bottom of the incline is 1.50 m/s. How far does the box travel along the incline before coming to rest?


Please help guys!
Thanks!

/prettynerd.
 

Answers and Replies

  • #2
Pyrrhus
Homework Helper
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Draw a Freebody Diagram for the cup in problem #2

Draw a Freebody Diagram for the box in problem #3, and use Kinematics equations for uniform acc. , the acceleration will be constant because it's produced by constant forces.
 
  • #3
Diane_
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For number 1, ask yourself - according to the mathematical model you're using, what causes friction? What force or forces does it depend on? And, will any of them be reduced by, say, pulling it as opposed to pushing it?
 
  • #4
El Hombre Invisible
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There's not enough information in #2 to answer it. If the coefficient is 0.30 N, then the force required to move it is:

F = 0.30 N x mg

so the maximum acceleration of the table is 0.30 N x g. However, since you haven't given the altitude of the plane, g is unknown (g = GM/r^2 where M is mass of Earth).

For #3, I suggest making the x-axis parallel to the incline and finding the perpendicular component of the object's weight to determine the normal reaction in turns of mg: N(y) = -mg sin theta. When you determine the acceleration acting on the object parallel to the incline (from F = 0.180 x N), the m will cancel out (a = F/m).
 
  • #5
HallsofIvy
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El Hombre Invisible said:
There's not enough information in #2 to answer it. If the coefficient is 0.30 N, then the force required to move it is:

F = 0.30 N x mg

so the maximum acceleration of the table is 0.30 N x g. However, since you haven't given the altitude of the plane, g is unknown (g = GM/r^2 where M is mass of Earth).

Oh, c'mon. How high would an airplane have to be for that to be significant?
 
  • #6
El Hombre Invisible
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HallsofIvy said:
Oh, c'mon. How high would an airplane have to be for that to be significant?
I don't know. I have no sense of perspective. I know g varies by around 0.03 m/s^2 just on the surface of the planet, maybe more. Maybe that's not significant. Most questions I've been set usually ask you to take g as 9.81 m/s^2 on ground level. In a plane, that might be off enough to be considered erroneous at that number of significant figures. If I was setting the question, I'd have defined either the altitude or g. But that's me. [silent pause] I'll get my coat.
 
  • #7
killerinstinct
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1. To answer what Diane has hinted toward...
If ou push a sled and theta > 0m there is a vertically downward component to you force. hence the normal force upward exerted by the ground will be larger than mg. If you pull the sled, your force has a vertically upword component, so the normal force Fn can be less than mg. Because friction force is proportional to the normal force, it will be less if you pull the sled.
 
  • #8
Diane_
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killerinstinct said:
1. To answer what Diane has hinted toward...
If ou push a sled and theta > 0m there is a vertically downward component to you force. hence the normal force upward exerted by the ground will be larger than mg. If you pull the sled, your force has a vertically upword component, so the normal force Fn can be less than mg. Because friction force is proportional to the normal force, it will be less if you pull the sled.

Umm... the reason I didn't come right out and say that is that the goal is sort of to get her to figure that out. She derives much less benefit from being told than from reasoning things out herself. Just for future information...
 

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