
#1
Sep212, 03:32 AM

PF Gold
P: 1,023

1. The problem statement, all variables and given/known data
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. 2. Relevant equations 3. 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(1cos60°) 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... 



#2
Sep212, 04:04 AM

P: 961

F is a conservative force.




#3
Sep212, 04:53 AM

PF Gold
P: 1,023

No... F is force applied by some external agent.




#4
Sep212, 04:55 AM

PF Gold
P: 1,023

Block moving on a circular trackWork energy Circular Motion problem
The block is pulled via a light inextensible string by external force F.




#5
Sep212, 05:12 AM

Mentor
P: 40,889





#6
Sep212, 05:18 AM

PF Gold
P: 1,023

Thanks Doc Al ...I have thought about displacement in the direction of constant force but not able to comprehend ...Sorry......please explain




#7
Sep212, 05:22 AM

PF Gold
P: 1,023

I feel the direction of force is continuously changing as the block moves.




#8
Sep212, 05:45 AM

Mentor
P: 40,889





#9
Sep212, 05:55 AM

HW Helper
Thanks
P: 9,818

ehild 



#10
Sep212, 06:44 AM

PF Gold
P: 1,023

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
Sep212, 06:51 AM

PF Gold
P: 1,023

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
Sep212, 06:56 AM

Mentor
P: 40,889





#13
Sep212, 06:58 AM

Mentor
P: 40,889

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
Sep212, 07:16 AM

PF Gold
P: 1,023

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
Sep212, 07:23 AM

Mentor
P: 40,889





#16
Sep212, 07:30 AM

PF Gold
P: 1,023

Displacement in the direction of constant force should be 10sin60° .Work done accordingly is 100 × 10(√3/2) ??? is this the work done ??




#17
Sep212, 07:34 AM

Mentor
P: 40,889

Edit: My mistake! Please use the displacement of the applied forcehow much the string movesnot the displacement of the block. 



#18
Sep212, 08:51 AM

HW Helper
Thanks
P: 9,818

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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