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

- anonymous

- jamiebookeater

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- anonymous

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

- anonymous

take the example of work done by friction force.

- anonymous

can you first explain the question ? :)

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## More answers

- anonymous

|dw:1348895881344:dw|

- anonymous

ok

- anonymous

there are several forces in the nature..name them...whatever u know.

- anonymous

like electric force, magnetic force, spring force, friction force, gravitational force...

- anonymous

right?

- anonymous

yeah right

- anonymous

|dw:1348896083024:dw|

- anonymous

and these are all the closed loops

- 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

reply*

- anonymous

|dw:1348896151349:dw|

- anonymous

now suppose there is a ball situated at point A near the earth

- anonymous

okay

- 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

okay

- anonymous

|dw:1348896421255:dw|

- anonymous

yes

- 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

look for the angle between the force and the displacement

- anonymous

90

- anonymous

cos 90 = 0

- anonymous

yes..so work done by gravitational force( A to B)=0

- anonymous

okay

- anonymous

our objective is the work done by the gravitational force on the ball
so now we come to B to C

- anonymous

|dw:1348896858314:dw|

- anonymous

ball is moving from B to C
again work done by the gravitational force = force*( displacement) *cos(theta)
theta=?

- anonymous

0 ?

- anonymous

i mean cos 0

- anonymous

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

- anonymous

displacement ( B to C) is vertically upward.

- anonymous

|dw:1348897246563:dw|

- anonymous

|dw:1348897411835:dw|

- anonymous

does it make sense for visualizing the angle between two vectors pictorially?

- anonymous

yes it does

- anonymous

so now we come to B to C|dw:1348897603389:dw|

- anonymous

got it?

- anonymous

force is vertically downward and displacement is vertically upward so angle between them is......?

- anonymous

okay cos 180

- anonymous

-1

- anonymous

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

- anonymous

yeah right

- anonymous

till now...it's fine?

- anonymous

yes :)

- anonymous

now work done by gravitational force (C to D)= 0
does it make sense?

- anonymous

yes it does

- anonymous

|dw:1348898023808:dw|

- 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

just look for angle between gravitational force and displacement when ball moves from D to A

- anonymous

cos 0 ?

- anonymous

yes

- anonymous

so work done by the gravitational force on the ball (D to A)= mg* h* cos(0)= mgh
BC=DA=h

- anonymous

okay

- 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

yes !

- anonymous

so total work done= W(A-->B) + W(B--->c) + W(C--->D) + W(D--->A)
=0 + (-mgh) + 0 + mgh
=0

- anonymous

so work done by gravitational force in the motion of a body over a closed loop is 0.

- anonymous

okay understood , and why is this not the same for every force in nature ?

- anonymous

but we took a rectangular loop for a closed loop..one can show this for any arbitrary|dw:1348898826574:dw| closed loop

- anonymous

but it's not true for every force in nature...e.g. friction force

- anonymous

and why ?

- 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

i don't understand , why not for frictional force ?

- anonymous

yes...I am coming there

- anonymous

okay :)

- anonymous

1st have a look...why not zero?|dw:1348899134273:dw|

- anonymous

suppose there is a block of mass m moves on a rectangular loop, and the loop is kept horizontally on the earth

- anonymous

like before , okay !

- anonymous

|dw:1348899311394:dw|

- 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

because during motion, angle between friction force and displacement will be 180'

- anonymous

am sorry but i did not understand the last part !

- 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

yes the same

- 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

90

- anonymous

|dw:1348899953137:dw|

- anonymous

|dw:1348900005470:dw|

- anonymous

see..if block moves from A to B then displacement of block =AB...right?

- anonymous

direction of displacement is to the right and the direction of friction force is to the left

- anonymous

so again there are two vectors friction force and displacement...look for the angle between them carfully

- 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

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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