anonymous
  • anonymous
Is it necessary that work done in the motion of a body over a closed loop is 0 for every force in nature ? why ?
Physics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
no...it's not true for every force in nature
anonymous
  • anonymous
take the example of work done by friction force.
anonymous
  • anonymous
can you first explain the question ? :)

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anonymous
  • anonymous
|dw:1348895881344:dw|
anonymous
  • anonymous
ok
anonymous
  • anonymous
there are several forces in the nature..name them...whatever u know.
anonymous
  • anonymous
like electric force, magnetic force, spring force, friction force, gravitational force...
anonymous
  • anonymous
right?
anonymous
  • anonymous
yeah right
anonymous
  • anonymous
|dw:1348896083024:dw|
anonymous
  • anonymous
and these are all the closed loops
anonymous
  • anonymous
i have to keep refreshing OS as its goin down , so my replays may seem late n am sry for that ! okay understood.
anonymous
  • anonymous
reply*
anonymous
  • anonymous
|dw:1348896151349:dw|
anonymous
  • anonymous
now suppose there is a ball situated at point A near the earth
anonymous
  • anonymous
okay
anonymous
  • anonymous
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
anonymous
  • anonymous
okay
anonymous
  • anonymous
|dw:1348896421255:dw|
anonymous
  • anonymous
yes
anonymous
  • anonymous
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)
anonymous
  • anonymous
look for the angle between the force and the displacement
anonymous
  • anonymous
90
anonymous
  • anonymous
cos 90 = 0
anonymous
  • anonymous
yes..so work done by gravitational force( A to B)=0
anonymous
  • anonymous
okay
anonymous
  • anonymous
our objective is the work done by the gravitational force on the ball so now we come to B to C
anonymous
  • anonymous
|dw:1348896858314:dw|
anonymous
  • anonymous
ball is moving from B to C again work done by the gravitational force = force*( displacement) *cos(theta) theta=?
anonymous
  • anonymous
0 ?
anonymous
  • anonymous
i mean cos 0
anonymous
  • anonymous
there are two vectors..force (mg, vertically downward) and displacement( B to C)..so think carefully for angle between them.
anonymous
  • anonymous
displacement ( B to C) is vertically upward.
anonymous
  • anonymous
|dw:1348897246563:dw|
anonymous
  • anonymous
|dw:1348897411835:dw|
anonymous
  • anonymous
does it make sense for visualizing the angle between two vectors pictorially?
anonymous
  • anonymous
yes it does
anonymous
  • anonymous
so now we come to B to C|dw:1348897603389:dw|
anonymous
  • anonymous
got it?
anonymous
  • anonymous
force is vertically downward and displacement is vertically upward so angle between them is......?
anonymous
  • anonymous
okay cos 180
anonymous
  • anonymous
-1
anonymous
  • anonymous
yes...right..:) so work done by the gravitational force = force*( displacement) *cos(theta) = mg * h * (-1) =-mgh suppose BC=h
anonymous
  • anonymous
yeah right
anonymous
  • anonymous
till now...it's fine?
anonymous
  • anonymous
yes :)
anonymous
  • anonymous
now work done by gravitational force (C to D)= 0 does it make sense?
anonymous
  • anonymous
yes it does
anonymous
  • anonymous
|dw:1348898023808:dw|
anonymous
  • anonymous
now we come to work done by the gravitational force on the ball (D to A) tell me how much it'll be?
anonymous
  • anonymous
just look for angle between gravitational force and displacement when ball moves from D to A
anonymous
  • anonymous
cos 0 ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
so work done by the gravitational force on the ball (D to A)= mg* h* cos(0)= mgh BC=DA=h
anonymous
  • anonymous
okay
anonymous
  • anonymous
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.
anonymous
  • anonymous
yes !
anonymous
  • anonymous
so total work done= W(A-->B) + W(B--->c) + W(C--->D) + W(D--->A) =0 + (-mgh) + 0 + mgh =0
anonymous
  • anonymous
so work done by gravitational force in the motion of a body over a closed loop is 0.
anonymous
  • anonymous
okay understood , and why is this not the same for every force in nature ?
anonymous
  • anonymous
but we took a rectangular loop for a closed loop..one can show this for any arbitrary|dw:1348898826574:dw| closed loop
anonymous
  • anonymous
but it's not true for every force in nature...e.g. friction force
anonymous
  • anonymous
and why ?
anonymous
  • anonymous
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
anonymous
  • anonymous
i don't understand , why not for frictional force ?
anonymous
  • anonymous
yes...I am coming there
anonymous
  • anonymous
okay :)
anonymous
  • anonymous
1st have a look...why not zero?|dw:1348899134273:dw|
anonymous
  • anonymous
suppose there is a block of mass m moves on a rectangular loop, and the loop is kept horizontally on the earth
anonymous
  • anonymous
like before , okay !
anonymous
  • anonymous
|dw:1348899311394:dw|
anonymous
  • anonymous
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?
anonymous
  • anonymous
because during motion, angle between friction force and displacement will be 180'
anonymous
  • anonymous
am sorry but i did not understand the last part !
anonymous
  • anonymous
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?
anonymous
  • anonymous
yes the same
anonymous
  • anonymous
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=?
anonymous
  • anonymous
90
anonymous
  • anonymous
|dw:1348899953137:dw|
anonymous
  • anonymous
|dw:1348900005470:dw|
anonymous
  • anonymous
see..if block moves from A to B then displacement of block =AB...right?
anonymous
  • anonymous
direction of displacement is to the right and the direction of friction force is to the left
anonymous
  • anonymous
so again there are two vectors friction force and displacement...look for the angle between them carfully
anonymous
  • anonymous
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?
anonymous
  • anonymous
Forces in nature can be either conservative or non-conservative. Work done in a closed loop by a conservative force is always 0 .

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