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ashna

Is it necessary that work done in the motion of a body over a closed loop is 0 for every force in nature ? why ?

  • one year ago
  • one year ago

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  1. akash123
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    no...it's not true for every force in nature

    • one year ago
  2. akash123
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    take the example of work done by friction force.

    • one year ago
  3. ashna
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    can you first explain the question ? :)

    • one year ago
  4. akash123
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    |dw:1348895881344:dw|

    • one year ago
  5. akash123
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    ok

    • one year ago
  6. akash123
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    there are several forces in the nature..name them...whatever u know.

    • one year ago
  7. akash123
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    like electric force, magnetic force, spring force, friction force, gravitational force...

    • one year ago
  8. akash123
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    right?

    • one year ago
  9. ashna
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    yeah right

    • one year ago
  10. akash123
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    |dw:1348896083024:dw|

    • one year ago
  11. akash123
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    and these are all the closed loops

    • one year ago
  12. ashna
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    i have to keep refreshing OS as its goin down , so my replays may seem late n am sry for that ! okay understood.

    • one year ago
  13. ashna
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    reply*

    • one year ago
  14. akash123
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    |dw:1348896151349:dw|

    • one year ago
  15. akash123
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    now suppose there is a ball situated at point A near the earth

    • one year ago
  16. ashna
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    okay

    • one year ago
  17. akash123
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    and the ball is taken from point A to B to C to D to A..that is ball has been moved on a closed loop during this motion

    • one year ago
  18. ashna
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    okay

    • one year ago
  19. akash123
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    |dw:1348896421255:dw|

    • one year ago
  20. ashna
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    yes

    • one year ago
  21. akash123
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    earth is exerting force mg on the ball...so tell me work done by this gravitational force if ball moves from point A to B...use work done= force* displacement * cos(theta)

    • one year ago
  22. akash123
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    look for the angle between the force and the displacement

    • one year ago
  23. ashna
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    90

    • one year ago
  24. ashna
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    cos 90 = 0

    • one year ago
  25. akash123
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    yes..so work done by gravitational force( A to B)=0

    • one year ago
  26. ashna
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    okay

    • one year ago
  27. akash123
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    our objective is the work done by the gravitational force on the ball so now we come to B to C

    • one year ago
  28. akash123
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    |dw:1348896858314:dw|

    • one year ago
  29. akash123
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    ball is moving from B to C again work done by the gravitational force = force*( displacement) *cos(theta) theta=?

    • one year ago
  30. ashna
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    0 ?

    • one year ago
  31. ashna
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    i mean cos 0

    • one year ago
  32. akash123
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    there are two vectors..force (mg, vertically downward) and displacement( B to C)..so think carefully for angle between them.

    • one year ago
  33. akash123
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    displacement ( B to C) is vertically upward.

    • one year ago
  34. akash123
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    |dw:1348897246563:dw|

    • one year ago
  35. akash123
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    |dw:1348897411835:dw|

    • one year ago
  36. akash123
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    does it make sense for visualizing the angle between two vectors pictorially?

    • one year ago
  37. ashna
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    yes it does

    • one year ago
  38. akash123
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    so now we come to B to C|dw:1348897603389:dw|

    • one year ago
  39. akash123
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    got it?

    • one year ago
  40. akash123
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    force is vertically downward and displacement is vertically upward so angle between them is......?

    • one year ago
  41. ashna
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    okay cos 180

    • one year ago
  42. ashna
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    -1

    • one year ago
  43. akash123
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    yes...right..:) so work done by the gravitational force = force*( displacement) *cos(theta) = mg * h * (-1) =-mgh suppose BC=h

    • one year ago
  44. ashna
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    yeah right

    • one year ago
  45. akash123
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    till now...it's fine?

    • one year ago
  46. ashna
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    yes :)

    • one year ago
  47. akash123
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    now work done by gravitational force (C to D)= 0 does it make sense?

    • one year ago
  48. ashna
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    yes it does

    • one year ago
  49. akash123
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    |dw:1348898023808:dw|

    • one year ago
  50. akash123
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    now we come to work done by the gravitational force on the ball (D to A) tell me how much it'll be?

    • one year ago
  51. akash123
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    just look for angle between gravitational force and displacement when ball moves from D to A

    • one year ago
  52. ashna
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    cos 0 ?

    • one year ago
  53. akash123
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    yes

    • one year ago
  54. akash123
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    so work done by the gravitational force on the ball (D to A)= mg* h* cos(0)= mgh BC=DA=h

    • one year ago
  55. ashna
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    okay

    • one year ago
  56. akash123
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    so our objective has been completed...I mean...we have calculated...work done by gravitational force on the ball when the ball moved on a closed loop.

    • one year ago
  57. ashna
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    yes !

    • one year ago
  58. akash123
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    so total work done= W(A-->B) + W(B--->c) + W(C--->D) + W(D--->A) =0 + (-mgh) + 0 + mgh =0

    • one year ago
  59. akash123
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    so work done by gravitational force in the motion of a body over a closed loop is 0.

    • one year ago
  60. ashna
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    okay understood , and why is this not the same for every force in nature ?

    • one year ago
  61. akash123
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    but we took a rectangular loop for a closed loop..one can show this for any arbitrary|dw:1348898826574:dw| closed loop

    • one year ago
  62. akash123
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    but it's not true for every force in nature...e.g. friction force

    • one year ago
  63. ashna
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    and why ?

    • one year ago
  64. akash123
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    so we classify the forces into two type if work done by the force over a closed loop is zero then conservative force otherwise non-conservative force

    • one year ago
  65. ashna
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    i don't understand , why not for frictional force ?

    • one year ago
  66. akash123
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    yes...I am coming there

    • one year ago
  67. ashna
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    okay :)

    • one year ago
  68. akash123
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    1st have a look...why not zero?|dw:1348899134273:dw|

    • one year ago
  69. akash123
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    suppose there is a block of mass m moves on a rectangular loop, and the loop is kept horizontally on the earth

    • one year ago
  70. ashna
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    like before , okay !

    • one year ago
  71. akash123
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    |dw:1348899311394:dw|

    • one year ago
  72. akash123
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    do u agree that work done by the friction force on the block will be negative in each segment(AB, BC,CD,DA) of this loop?

    • one year ago
  73. akash123
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    because during motion, angle between friction force and displacement will be 180'

    • one year ago
  74. ashna
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    am sorry but i did not understand the last part !

    • one year ago
  75. akash123
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    which one...do u agree that work done by the friction force on the block will be negative in each segment(AB, BC,CD,DA) of this loop?

    • one year ago
  76. ashna
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    yes the same

    • one year ago
  77. akash123
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    ok...1st we'll see the work done by the friction force on the block,when it moves from A to B again work done= force*( displacement)* cos(theta) theta=?

    • one year ago
  78. ashna
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    90

    • one year ago
  79. akash123
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    |dw:1348899953137:dw|

    • one year ago
  80. akash123
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    |dw:1348900005470:dw|

    • one year ago
  81. akash123
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    see..if block moves from A to B then displacement of block =AB...right?

    • one year ago
  82. akash123
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    direction of displacement is to the right and the direction of friction force is to the left

    • one year ago
  83. akash123
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    so again there are two vectors friction force and displacement...look for the angle between them carfully

    • one year ago
  84. sam291
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    But when it moves from C to D the direction of the force vector is opposite from the direction from A to B, so it will still add up to zero going around the loop. Am I wrong?

    • one year ago
  85. kappa007
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    Forces in nature can be either conservative or non-conservative. Work done in a closed loop by a conservative force is always 0 .

    • one year ago
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