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

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?

$$\mathrm{A}\:\mathrm{force}\:\overset{\rightarrow} {{F}}\:=\:\mathrm{2}{x}\overset{\wedge} {{j}}\:\mathrm{newton}\:\mathrm{acts}\:\mathrm{in}\:\mathrm{a}\:\mathrm{region} \\ $$$$\mathrm{where}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{anticlockwise} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{square}\:\mathrm{loop}\:\mathrm{of}\:\mathrm{2}\:\mathrm{m}\:\mathrm{in}\:{x}-{y}\:\mathrm{plane}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{total}\:\mathrm{amount}\:\mathrm{of}\:\mathrm{work} \\ $$$$\mathrm{done}.\:\mathrm{Is}\:\mathrm{this}\:\mathrm{force}\:\mathrm{a}\:\mathrm{conservative}\:\mathrm{force} \\ $$$$\mathrm{or}\:\mathrm{a}\:\mathrm{non}-\mathrm{conservative}\:\mathrm{force}? \\ $$

Commented by ajfour last updated on 22/Oct/17

what if F =F_0 (xi^� +yj^� ) ?

$${what}\:{if}\:{F}\:={F}_{\mathrm{0}} \left({x}\hat {{i}}+{y}\hat {{j}}\right)\:? \\ $$

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

$$\mathrm{then}\:\mathrm{it}\:\mathrm{would}\:\mathrm{be}\:\mathrm{a}\:\mathrm{conservative} \\ $$$$\mathrm{force}\:\mathrm{and}\:\mathrm{work}\:\mathrm{done}\:\mathrm{would} \\ $$$$\mathrm{be}\:\mathrm{zero}\:\mathrm{joules}\:\mathrm{from}\:\mathrm{A}\:\mathrm{to}\:\mathrm{A} \\ $$$$ \\ $$

Commented by ajfour last updated on 22/Oct/17

W= = ∫F_0 (xi^� +yj^� ).(dxi^� +dyj^� ) W=∫F^� .ds^� = F_0 ∫(xdx+ydy) =(F_0 /2)[(x^2 −x_0 ^2 )+(y^2 −y_0 ^2 )]. while if F=2xj^� , then W=∫(2xj^� ).ds^� =∫_(ABCDA) 2xdy =(0+8+0+0)J =8J .

$${W}=\:=\:\int{F}_{\mathrm{0}} \left({x}\hat {{i}}+{y}\hat {{j}}\right).\left({dx}\bar {{i}}+{dy}\hat {{j}}\right) \\ $$$${W}=\int\bar {{F}}.{d}\bar {{s}}\:=\:{F}_{\mathrm{0}} \int\left({xdx}+{ydy}\right) \\ $$$$\:\:\:\:=\frac{{F}_{\mathrm{0}} }{\mathrm{2}}\left[\left({x}^{\mathrm{2}} −{x}_{\mathrm{0}} ^{\mathrm{2}} \right)+\left({y}^{\mathrm{2}} −{y}_{\mathrm{0}} ^{\mathrm{2}} \right)\right]. \\ $$$${while}\:{if}\:{F}=\mathrm{2}{x}\hat {{j}}\:,\:{then} \\ $$$$\:\:{W}=\int\left(\mathrm{2}{x}\hat {{j}}\right).{d}\bar {{s}}\:=\int_{{ABCDA}} \mathrm{2}{xdy} \\ $$$$=\left(\mathrm{0}+\mathrm{8}+\mathrm{0}+\mathrm{0}\right){J}\:=\mathrm{8}\boldsymbol{{J}}\:. \\ $$$$ \\ $$

Commented by Tinkutara last updated on 22/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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

Work done from A to B : since the direction of force is ⊥ to the displacement ⇒ W = 0 J Work done from B to C : F = 2×2 = 4 and is parallel to displacement ⇒ W = F^→ ∙d^→ = (4×2 )×Cos 0° = 8 J From C to D : W = 0 as force is ⊥ to displacement From D to A : W = 0 as F = 0. Total work done_(from A to A) = 8 J force is non conservative

$$ \\ $$$${Work}\:{done}\:{from}\:{A}\:{to}\:{B}\:: \\ $$$${since}\:{the}\:{direction}\:{of}\:{force} \\ $$$${is}\:\bot\:{to}\:{the}\:{displacement} \\ $$$$\Rightarrow\:{W}\:=\:\mathrm{0}\:{J} \\ $$$${Work}\:{done}\:{from}\:{B}\:{to}\:{C}\:: \\ $$$${F}\:=\:\mathrm{2}×\mathrm{2}\:=\:\mathrm{4}\:{and}\:{is}\:{parallel} \\ $$$${to}\:{displacement} \\ $$$$\Rightarrow\:{W}\:=\:\overset{\rightarrow} {{F}}\:\centerdot\overset{\rightarrow} {{d}}\:=\:\left(\mathrm{4}×\mathrm{2}\:\right)×{Cos}\:\mathrm{0}° \\ $$$$=\:\mathrm{8}\:{J} \\ $$$${From}\:{C}\:{to}\:{D}\::\: \\ $$$${W}\:=\:\mathrm{0}\:{as}\:{force}\:{is}\:\bot\:{to}\: \\ $$$${displacement} \\ $$$${From}\:{D}\:{to}\:{A}\:: \\ $$$${W}\:=\:\mathrm{0}\:{as}\:{F}\:=\:\mathrm{0}. \\ $$$$ \\ $$$${Total}\:{work}\:{done}_{{from}\:{A}\:{to}\:{A}} \:=\:\mathrm{8}\:{J} \\ $$$$\:{force}\:{is}\:{non}\:{conservative} \\ $$

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.

$$\mathrm{Work}\:\mathrm{done}\:\mathrm{from}\:\mathrm{B}\:\mathrm{to}\:\mathrm{C}\:\mathrm{is}\:\mathrm{why}\:\mathrm{not}\:\mathrm{0}? \\ $$$$\mathrm{Since}\:\mathrm{there}\:{x}\:\mathrm{does}\:\mathrm{not}\:\mathrm{change}. \\ $$

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.

$$\mathrm{because}\:\mathrm{it}\:\mathrm{is}\:\mathrm{displacing}\:\mathrm{in} \\ $$$$\mathrm{y}\:\mathrm{direction}\:.\:\mathrm{And}\:\mathrm{force}\:\mathrm{is} \\ $$$$\mathrm{eing}\:\mathrm{constant}\:\:\mathrm{as}\:\mathrm{x}\:\mathrm{constant}. \\ $$

Commented by Tinkutara last updated on 22/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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

you are welcome.

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}. \\ $$

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