Question
Given ⃗F=(xy2)^i+(x2y)^jN. The work done by ⃗F when a particle is taken along the semicircular path OAB where the coordinates of B are (4,0) is
Answer: Option D
:
D
Given⃗F=(xy2)^i+(x2y)^j
W = ∫Fxdx+∫Fydy
= ∫xy2dx+∫x2ydy
= 12∫d(x2y2)=[x2y22](4,0)(0,0)=0
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:
D
Given⃗F=(xy2)^i+(x2y)^j
W = ∫Fxdx+∫Fydy
= ∫xy2dx+∫x2ydy
= 12∫d(x2y2)=[x2y22](4,0)(0,0)=0
Was this answer helpful ?
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