A force F^→ = 2xj^∧ newton acts in a region where a particle moves anticlockwise in a square loop of 2 m in x-y plane. Calculate the total amount of work done. Is this force a conservative force or a non-conservative force?


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Question Number 22696 by Tinkutara last updated on 22/Oct/17

$A force \vec{F} = 2 x \overset{\land}{j} newton acts in a region$ $where a particle moves anticlockwise$ $in a square loop of 2 m in x - y plane .$ $Calculate the total amount of work$ $done . Is this force a conservative force$ $or a non - conservative force ?$
Commented by ajfour last updated on 22/Oct/17

$w h a t i f F = F_{0} (x \hat{i} + y \hat{j}) ?$
Commented by Tinkutara last updated on 22/Oct/17

Commented by Sahib singh last updated on 22/Oct/17

$then it would be a conservative$ $force and work done would$ $be zero joules from A to A$
Commented by ajfour last updated on 22/Oct/17

$W = = \int F_{0} (x \hat{i} + y \hat{j}) . (d x \bar{i} + d y \hat{j})$ $W = \int \bar{F} . d \bar{s} = F_{0} \int (x d x + y d y)$ $= \frac{F_{0}}{2} [(x^{2} - x_{0}^{2}) + (y^{2} - y_{0}^{2})] .$ $w h i l e i f F = 2 x \hat{j}, t h e n$ $W = \int (2 x \hat{j}) . d \bar{s} = \int_{A B C D A} 2 x d y$ $= (0 + 8 + 0 + 0) J = 8 J .$
Commented by Tinkutara last updated on 22/Oct/17

$Thank you very much Sir!$
	Answered by Sahib singh last updated on 22/Oct/17

	$W o r k d o n e f r o m A t o B :$ $s i n c e t h e d i r e c t i o n o f f o r c e$ $i s ⊥ t o t h e d i s p l a c e m e n t$ $\Rightarrow W = 0 J$ $W o r k d o n e f r o m B t o C :$ $F = 2 \times 2 = 4 a n d i s p a r a l l e l$ $t o d i s p l a c e m e n t$ $\Rightarrow W = \vec{F} \cdot \vec{d} = (4 \times 2) \times C o s 0 °$ $= 8 J$ $F r o m C t o D :$ $W = 0 a s f o r c e i s ⊥ t o$ $d i s p l a c e m e n t$ $F r o m D t o A :$ $W = 0 a s F = 0 .$ $T o t a l w o r k {d o n e}_{f r o m A t o A} = 8 J$ $f o r c e i s n o n c o n s e r v a t i v e$
	Commented by Tinkutara last updated on 22/Oct/17

	$Work done from B to C is why not 0 ?$ $Since there x does not change .$
	Commented by Sahib singh last updated on 22/Oct/17

	$because it is displacing in$ $y direction . And force is$ $eing constant as x constant .$
	Commented by Tinkutara last updated on 22/Oct/17

	$Thank you very much Sir!$
	Commented by Sahib singh last updated on 22/Oct/17

	$you are welcome .$
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