Block On Incline w/ Tension

  • Thread starter venceslau
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  • #1
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Homework Statement


A mass m = 16.0 kg is pulled along a horizontal floor, with a coefficient of kinetic friction μk = 0.14, for a distance d = 8.3 m. Then the mass is continued to be pulled up a frictionless incline that makes an angle θ = 38.0° with the horizontal. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of θ = 38.0° (thus on the incline it is parallel to the surface) and has a tension T = 56.0 N.


2. The attempt at a solution
1) What is the work done by tension before the block gets to the incline?

T*cos(rad(theta))*d = 366.267398276404 J

2) What is the work done by friction as the block slides on the flat horizontal surface?

-muk*m*g*d = -182.38752 J

3) What is the speed of the block right before it begins to travel up the incline?

sqrt(2*(366.267+(-182.38))/m) = 4.79435866409679 m/s

4) How far up the incline does the block travel before coming to rest?
I have tried everything I can think of, but I keep getting it wrong...

5) What is the work done by gravity as it comes to rest?
This question depends on 4 if I'm not mistaken, so haven't been able to do it yet...
 

Answers and Replies

  • #2
Doc Al
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1) What is the work done by tension before the block gets to the incline?

T*cos(rad(theta))*d = 366.267398276404 J
OK.

2) What is the work done by friction as the block slides on the flat horizontal surface?

-muk*m*g*d = -182.38752 J
Careful. Friction = μN. Since the rope is pulling up on the block, the normal force is not simply mg.

3) What is the speed of the block right before it begins to travel up the incline?

sqrt(2*(366.267+(-182.38))/m) = 4.79435866409679 m/s
You'll need to redo this one.

4) How far up the incline does the block travel before coming to rest?
I have tried everything I can think of, but I keep getting it wrong...
What's the acceleration of the block when it's on the incline?
 
  • #3
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Doc Al, thanks for your prompt reply!

This is an online system, and it only took the answer as being correct once I excluded the Y component of the Tension. But I agree with you, it should be factored in there. All the equations I presented above, have already been accepted by the system.

Acceleration of the block on the incline:
-(g*sin(theta) - T/m) ...?

Edit: I finally got it, thanks for the push Doc Al, that was the question I needed asked! :D
 
Last edited:

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