Block moving on a circular track-Work energy Circular Motion problem

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Homework Statement



A block of mass 1 Kg initially at rest at point B is being pulled by a constant force F= 100N along a smooth circular track of radius 10 m . Find the velocity of the block when it reaches point P making an angle 60 with the vertical.



Homework Equations





The Attempt at a Solution



KE1+PE1+W[F]=KE2+PE2

Now KE1=PE1=0 assuming reference level to be at B.
KE2=1/2(mv[2]
PE2=mgl(1-cos60°)
W[F]=?

How do we calculate work done by force F? My understanding says that angle between constant force and displacement changes continuously as the block moves along the track.Please help me.Thanks...
 

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Answers and Replies

  • #2
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F is a conservative force.
 
  • #3
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No... F is force applied by some external agent.
 
  • #4
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The block is pulled via a light inextensible string by external force F.
 
  • #5
Doc Al
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How do we calculate work done by force F? My understanding says that angle between constant force and displacement changes continuously as the block moves along the track.
Yes, the angle continually changes but luckily the force is constant. What's the displacement in the direction of that constant force?
 
  • #6
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Thanks Doc Al ...I have thought about displacement in the direction of constant force but not able to comprehend ...Sorry......please explain
 
  • #7
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I feel the direction of force is continuously changing as the block moves.
 
  • #8
Doc Al
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I feel the direction of force is continuously changing as the block moves.
So it's not a constant force? Did you misstate something in the problem statement?
 
  • #9
ehild
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Homework Statement



A block of mass 1 Kg initially at rest at point B is being pulled by a constant force F= 100N along a smooth circular track of radius 10 m .
Tanya, is the magnitude of the force constant and acting along the track or is it really a constant horizontal force? The drawing suggest a constant horizontal force.

ehild
 
  • #10
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The block is being pulled by a light inextensible string which passes over a small frictionless massless pulley . The string is pulled by a constant horizontal force of magnitude 100 N .
 
  • #11
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What i meant by change of direction of force is that the magnitude of tension in the string pulling the block will remain constant but the direction of this tension force changes.Whereas the displacement of the block is always tangential to the circular track.
 
  • #12
Doc Al
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The string is pulled by a constant horizontal force of magnitude 100 N .
Good. So the direction of that force F is kept horizontal. (Makes things easier.)
 
  • #13
Doc Al
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What i meant by change of direction of force is that the magnitude of tension in the string pulling the block will remain constant but the direction of this tension force changes.Whereas the displacement of the block is always tangential to the circular track.
Does the tension do any work on the block?

As far as the work done by that constant force F, go back to my earlier question: What's the displacement in the direction of that constant force?
 
  • #14
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Please correct me if i am wrong...the constant force is being applied on the string which in turn pulls the block.So , now external force on the block is the tension .Now since the string is massless the force applied on the string = Tension in the string. But the angle between this tension force and the displacement of the block changes continuously isnt it ??

Doc Al ...We have to relate displacement of the block with the tension not the constant horizontal force... isnt it ?
 
  • #15
Doc Al
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Please correct me if i am wrong...the constant force is being applied on the string which in turn pulls the block.So , now external force on the block is the tension .Now since the string is massless the force applied on the string = Tension in the string. But the angle between this tension force and the displacement of the block changes continuously isnt it ??
Yes, you are correct. I guess I wasn't paying close enough attention to your diagram. (I was thinking of something else. :redface:) Nonetheless, there's a much easier way to solve this and my suggestion stands as is.
Doc Al ...We have to relate displacement of the block with the tension not the constant horizontal force... isnt it ?
Ask yourself: How much work is done on the system by that constant force? That's all you need to know.
 
  • #16
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Displacement in the direction of constant force should be 10sin60° .Work done accordingly is 100 × 10(√3/2) ??? is this the work done ??
 
  • #17
Doc Al
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Displacement in the direction of constant force should be 10sin60° .Work done accordingly is 100 × 10(√3/2) ??? is this the work done ??
Exactly. Whatever is exerting that constant force is doing work on the system.

Edit: My mistake! Please use the displacement of the applied force--how much the string moves--not the displacement of the block.
 
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  • #18
ehild
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If the work is equal to the product of the force acting on the rope and the displacement of the body in the direction of that force, what is the work done on the block in the picture by the force F when it raises with height h? Is it zero because the force and displacement enclose 90°angle?

ehild
 

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  • #19
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echild... sir... please help me with the problem...What you have asked ...according to Doc Al... that should be zero...But i feel positive work is done when the block is raised because force acting on block is tension not F.Yes F=T. But It is T which is acting on the block not F.... Please help with my problem...
 
  • #20
ehild
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The tension acts along a chord which makes angle β with the tangent of the circle. The work during displacement ds along the arc is Tds cos(β). β=(\pi/3-φ)/2. The total work of the rope is [tex]W=\int_0^{\pi/3}{TR\cos(\pi/6-0.5 \phi)d\phi}[/tex]

There is an easier approach: The rope is pulled by the horizontal constant force F. The end of the rope moves horizontally along a distance L, the work done by the pulling force is FL. The rope is inextensible, so L is equal to the length of the chord between the initial and final position of the block.

ehild
 

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  • #21
ehild
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echild... sir... please help me with the problem...What you have asked ...according to Doc Al... that should be zero...But i feel positive work is done when the block is raised because force acting on block is tension not F.Yes F=T. But It is T which is acting on the block not F.... Please help with my problem...
You are exactly right. See my next post.

ehild
 
  • #22
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brilliant...echild...this was what i was looking for :smile:....how have you calculated β ???
 
  • #23
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sir...how have you calculated β in terms of ∅ ???
 
  • #25
Doc Al
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echild... sir... please help me with the problem...What you have asked ...according to Doc Al... that should be zero...
Not at all. The force applied times the displacement of the point of application of that force is certainly not zero.
But i feel positive work is done when the block is raised because force acting on block is tension not F.Yes F=T. But It is T which is acting on the block not F.... Please help with my problem...
It's an excellent exercise to think in terms of the tension and compute the work done by that force directly. But there's an easier way.
 

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