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no...it's not true for every force in nature
take the example of work done by friction force.
can you first explain the question ? :)
there are several forces in the nature..name them...whatever u know.
like electric force, magnetic force, spring force, friction force, gravitational force...
and these are all the closed loops
i have to keep refreshing OS as its goin down , so my replays may seem late n am sry for that ! okay understood.
now suppose there is a ball situated at point A near the earth
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
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)
look for the angle between the force and the displacement
cos 90 = 0
yes..so work done by gravitational force( A to B)=0
our objective is the work done by the gravitational force on the ball so now we come to B to C
ball is moving from B to C again work done by the gravitational force = force*( displacement) *cos(theta) theta=?
i mean cos 0
there are two vectors..force (mg, vertically downward) and displacement( B to C)..so think carefully for angle between them.
displacement ( B to C) is vertically upward.
does it make sense for visualizing the angle between two vectors pictorially?
yes it does
so now we come to B to C|dw:1348897603389:dw|
force is vertically downward and displacement is vertically upward so angle between them is......?
okay cos 180
yes...right..:) so work done by the gravitational force = force*( displacement) *cos(theta) = mg * h * (-1) =-mgh suppose BC=h
till now...it's fine?
now work done by gravitational force (C to D)= 0 does it make sense?
yes it does
now we come to work done by the gravitational force on the ball (D to A) tell me how much it'll be?
just look for angle between gravitational force and displacement when ball moves from D to A
cos 0 ?
so work done by the gravitational force on the ball (D to A)= mg* h* cos(0)= mgh BC=DA=h
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.
so total work done= W(A-->B) + W(B--->c) + W(C--->D) + W(D--->A) =0 + (-mgh) + 0 + mgh =0
so work done by gravitational force in the motion of a body over a closed loop is 0.
okay understood , and why is this not the same for every force in nature ?
but we took a rectangular loop for a closed loop..one can show this for any arbitrary|dw:1348898826574:dw| closed loop
but it's not true for every force in nature...e.g. friction force
and why ?
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
i don't understand , why not for frictional force ?
yes...I am coming there
1st have a look...why not zero?|dw:1348899134273:dw|
suppose there is a block of mass m moves on a rectangular loop, and the loop is kept horizontally on the earth
like before , okay !
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?
because during motion, angle between friction force and displacement will be 180'
am sorry but i did not understand the last part !
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?
yes the same
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=?
see..if block moves from A to B then displacement of block =AB...right?
direction of displacement is to the right and the direction of friction force is to the left
so again there are two vectors friction force and displacement...look for the angle between them carfully
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?
Forces in nature can be either conservative or non-conservative. Work done in a closed loop by a conservative force is always 0 .