Find the work done in moving a particle once around a circle C in xy-plane,if the circle has centre at origin and radius 2 and if the force field is F = (2x-y+2z)i +(x+y-z)j+(3x-2y-5z)k.
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1
Answer:
workdone=0
because in circular motion workdone is always zero.
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Given:
Circle of radius , r =2
Force Field , F = (2x-y+2z)i +(x+y-z)j+(3x-2y-5z)k.
To find:
Find the work done in moving a particle once around a circle C in xy-plane.
Solution:
As We know that the workdone in closed path = ∫ F.dr
F = (2x-y+2z)i +(x+y-z)j+(3x-2y-5z)k.
dr = dxi + dyj ( dz= 0, x-y plane)
∫ F.dr = ∫{ (2x-y+2z)i +(x+y-z)j+(3x-2y-5z)k}. {dxi + dyj}
= ∫ { (2x-y+2z)dx} + ∫{(x+y-z) dj}
x= rcos t , y = rsint , z=0 t varies from 0 to 2π
W = ∫ { (4cost-2sint)dt} + ∫{(2cost+2sint) dt}
= ∫6cost dt
=6 ( sinx )
putting 0 and 2π
= 6 { sin (2π) - sin 0)}
= 0
Hence the workdone will be 0
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