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

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SUMMARY

The discussion revolves around calculating the velocity of a block with a mass of 1 Kg being pulled by a constant force of 100 N along a smooth circular track with a radius of 10 m. The key equation used is the work-energy principle, expressed as KE1 + PE1 + W[F] = KE2 + PE2. The participants clarify that while the force is constant, the direction of the force changes as the block moves, affecting the calculation of work done. The work done by the constant force is determined to be 100 × 10(√3/2), which is the product of the force and the displacement in the direction of the force.

PREREQUISITES
  • Understanding of the work-energy principle in physics
  • Familiarity with circular motion concepts
  • Knowledge of gravitational potential energy calculations
  • Ability to apply trigonometric functions in physics problems
NEXT STEPS
  • Study the work-energy theorem in detail
  • Learn about gravitational potential energy and its applications
  • Explore the dynamics of circular motion and tension in strings
  • Investigate the relationship between force, displacement, and work done
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of the work-energy principle in circular motion scenarios.

  • #31
Tanya Sharma said:
echild...i perfectly understand the first method...regarding second method...work done by pulling force is FL which is work done on the system...but system comprises of block and string wherein string is massless so its mechanical energy is zero .Total energy of system is equal to total energy of block ...?am i right ...this means if the string were having mass then we would not be able to apply the second method?because here

When you calculate the speed of the block, do not forget the work of gravity. The change of KE is equal to the work of all forces...

ehild
 
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  • #32
echild ...thanks a ton... what an explanation...u have made physics enjoyable...i m very very grateful:smile:

thanks to Doc Al ...u were correct but the language confused me...
 
  • #33
Thanks Tania. You are really very kind. I am pleased that you feel Physics enjoyable. It is.
Remember Doc Al sentence: The work done on a system by a force is
The force applied times the displacement of the point of application of that force
. It is very important.

ehild
 
  • #34
ehild said:
Well, have you got that coffee since then? :smile: You totally confused me, so as I deleted my post and had to rewrite it when the example of the simple pulley occurred to me. And I was surprised by the result of the integral method. (I blame it to the hot damp weather here. It is difficult to think.)

ehild
Sorry about that, ehild! And yes, I've been drinking a lot of coffee since then. :-p
 
  • #35
Doc Al said:
Sorry about that, ehild! And yes, I've been drinking a lot of coffee since then. :-p

Take care, too much coffee is bad to hearth. (I also had a coffee since then. :biggrin: )

ehild
 

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