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 ?
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akash123
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no...it's not true for every force in nature
akash123
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take the example of work done by friction force.
ashna
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can you first explain the question ? :)
akash123
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|dw:1348895881344:dw|
akash123
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ok
akash123
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there are several forces in the nature..name them...whatever u know.
akash123
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like electric force, magnetic force, spring force, friction force, gravitational force...
akash123
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right?
ashna
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yeah right
akash123
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|dw:1348896083024:dw|
akash123
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and these are all the closed loops
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.
ashna
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reply*
akash123
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|dw:1348896151349:dw|
akash123
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now suppose there is a ball situated at point A near the earth
ashna
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okay
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
ashna
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okay
akash123
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|dw:1348896421255:dw|
ashna
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yes
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)
akash123
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look for the angle between the force and the displacement
ashna
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90
ashna
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cos 90 = 0
akash123
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yes..so work done by gravitational force( A to B)=0
ashna
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okay
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
akash123
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|dw:1348896858314:dw|
akash123
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ball is moving from B to C
again work done by the gravitational force = force*( displacement) *cos(theta)
theta=?
ashna
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0 ?
ashna
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i mean cos 0
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.
akash123
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displacement ( B to C) is vertically upward.
akash123
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|dw:1348897246563:dw|
akash123
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|dw:1348897411835:dw|
akash123
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does it make sense for visualizing the angle between two vectors pictorially?
ashna
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yes it does
akash123
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so now we come to B to C|dw:1348897603389:dw|
akash123
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got it?
akash123
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force is vertically downward and displacement is vertically upward so angle between them is......?
ashna
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okay cos 180
ashna
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-1
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
ashna
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yeah right
akash123
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till now...it's fine?
ashna
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yes :)
akash123
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now work done by gravitational force (C to D)= 0
does it make sense?
ashna
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yes it does
akash123
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|dw:1348898023808:dw|
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?
akash123
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just look for angle between gravitational force and displacement when ball moves from D to A
ashna
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cos 0 ?
akash123
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yes
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
ashna
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okay
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.
ashna
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yes !
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
akash123
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so work done by gravitational force in the motion of a body over a closed loop is 0.
ashna
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okay understood , and why is this not the same for every force in nature ?
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
akash123
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but it's not true for every force in nature...e.g. friction force
ashna
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and why ?
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
ashna
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i don't understand , why not for frictional force ?
akash123
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yes...I am coming there
ashna
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okay :)
akash123
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1st have a look...why not zero?|dw:1348899134273:dw|
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
ashna
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like before , okay !
akash123
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|dw:1348899311394:dw|
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?
akash123
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because during motion, angle between friction force and displacement will be 180'
ashna
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am sorry but i did not understand the last part !
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?
ashna
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yes the same
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=?
ashna
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90
akash123
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|dw:1348899953137:dw|
akash123
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|dw:1348900005470:dw|
akash123
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see..if block moves from A to B then displacement of block =AB...right?
akash123
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direction of displacement is to the right and the direction of friction force is to the left
akash123
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so again there are two vectors friction force and displacement...look for the angle between them carfully
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?
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 .