A particle is moved along a path ABCDEFA, as shown in figure, in presence of a force vecF=(α y hati + 2 α x hatÈ·)N , Where x and y are in meter and α =-1 Nm- 1 . The work done on the particle by this force vecF ( textin J) is (p/100) , find p .

Question

A particle is moved along a path ABCDEFA, as shown in figure, in presence of a force vecF=(α y hati + 2 α x hatÈ·)N , Where x and y are in meter and α =-1 Nm- 1 . The work done on the particle by this force vecF ( textin J) is (p/100) , find p . <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-2vpazdzniz7rfyu4.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

As α =-1 ∴ vec F=- underset1 underbracey hati - x hatj-x hatj This is now a perfect differential format whose work done is zero for a complete cycle. Hence for -xj only WD needs to be calculated. ∴ W=1× 0.5+0.5× 0.5 =0.5+0.25 =0.75 J

Q. A particle is moved along a path $A BC D EF A,$ as shown in figure, in presence of a force $F = (α y \hat{i} + 2 αx \overset{ȷ}{^}) N$ , Where $x$ and $y$ are in meter and $α = - 1 N m^{- 1}$ . The work done on the particle by this force $F$ $(in J)$ is $\frac{p}{100}$ , find $p$ .

As $α = - 1$
$∴ F = - 1 y \hat{i} - x \hat{j} - x \hat{j}$
This is now a perfect differential format whose work done is zero for a complete cycle. Hence for $- x j$ only WD needs to be calculated.
$∴ W = 1 \times 0.5 + 0.5 \times 0.5$
$= 0.5 + 0.25$
$= 0.75 J$

Q. A particle is moved along a path ABCDEFA, as shown in figure, in presence of a force F=(αyi^+2αxȷ^​)N , Where x and y are in meter and α=−1Nm−1 . The work done on the particle by this force F (inJ) is 100p​ , find p .

As α=−1 ∴F=−1yi^−xj^​​​−xj^​ This is now a perfect differential format whose work done is zero for a complete cycle. Hence for −xj only WD needs to be calculated. ∴W=1×0.5+0.5×0.5 =0.5+0.25 =0.75J

Q. A particle is moved along a path $A BC D EF A,$ as shown in figure, in presence of a force $F = (α y \hat{i} + 2 αx \overset{ȷ}{^}) N$ , Where $x$ and $y$ are in meter and $α = - 1 N m^{- 1}$ . The work done on the particle by this force $F$ $(in J)$ is $\frac{p}{100}$ , find $p$ .

As $α = - 1$
$∴ F = - 1 y \hat{i} - x \hat{j} - x \hat{j}$
This is now a perfect differential format whose work done is zero for a complete cycle. Hence for $- x j$ only WD needs to be calculated.
$∴ W = 1 \times 0.5 + 0.5 \times 0.5$
$= 0.5 + 0.25$
$= 0.75 J$