A particle is moved along a path A-B-C-D-E-F- | vedclass

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(C) The work done is given by $W = \int \vec{F} \cdot d\vec{r} = \int (\alpha y dx + 2 \alpha x dy)$. Given $\alpha = -1$,$W = \int (-y dx - 2x dy)$.A

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(C) The work done is given by $W = \int \vec{F} \cdot d\vec{r} = \int (\alpha y dx + 2 \alpha x dy)$. Given $\alpha = -1$,$W = \int (-y dx - 2x dy)$.Along $AB$: $y=1, dy=0, x: 0 	o 1$. $W_{AB} = \int_0^1 (-1) dx = -1 \ J$.Along $BC$: $x=1, dx=0, y: 1 	o 0.5$. $W_{BC} = \int_1^{0.5} -2(1) dy = -2(0.5 - 1) = 1 \ J$.Along $CD$: $y=0.5, dy=0, x: 1 	o 0.5$. $W_{CD} = \int_1^{0.5} -0.5 dx = -0.5(-0.5) = 0.25 \ J$.Along $DE$: $x=0.5, dx=0, y: 0.5 	o 0$. $W_{DE} = \int_{0.5}^0 -2(0.5) dy = -1(-0.5) = 0.5 \ J$.Along $EF$: $y=0, dy=0, x: 0.5 	o 0$. $W_{EF} = \int_{0.5}^0 0 dx = 0 \ J$.Along $FA$: $x=0, dx=0, y: 0 	o 1$. $W_{FA} = \int_0^1 -2(0) dy = 0 \ J$.Total work $W = W_{AB} + W_{BC} + W_{CD} + W_{DE} + W_{EF} + W_{FA} = -1 + 1 + 0.25 + 0.5 + 0 + 0 = 0.75 \ J$.

$A$ particle is moved along a path $A-B-C-D-E-F-A$,as shown in the figure,in the presence of a force $\vec{F} = (\alpha y \hat{i} + 2 \alpha x \hat{j}) \ N$,where $x$ and $y$ are in meters and $\alpha = -1 \ N/m$. The work done on the particle by this force $\vec{F}$ will be . . . . . . Joule.

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