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PhysicsNumerical+3Question 13 of 54

JEE Advanced 2019 · Paper 1, Physics question 13

A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in figure, in presence of a force F⃗=(αyi^+2αxj^)\vec F = (\alpha y\hat i + 2\alpha x\hat j) N, where xx and yy are in meter and α=−1 Nm−1\alpha = -1\ \mathrm{Nm^{-1}}. The work done on the particle by this force F⃗\vec F will be ______ Joule.

A closed path in the x-y plane made of straight segments: A (0, 1) → B (1, 1) → C (1, 0.5) → D (0.5, 0.5) → E (0.5, 0) → F (0, 0) → A (0, 1).
Answer and worked solution

Official answer: 0.75

Use Green's theorem around the closed path:
W=∮(Fx dx+Fy dy)=∬(∂Fy∂x−∂Fx∂y)dA=(2α−α)×(signed area).W=\oint(F_x\,dx+F_y\,dy)=\iint\left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)dA=(2\alpha-\alpha)\times(\text{signed area}).

Signed area. The region is the unit square minus the 0.5×0.50.5\times0.5 corner square, area 0.750.75. The path A→B→C→D→E→F→A runs clockwise, so the signed area is −0.75-0.75.

Work.
W=α(−0.75)=(−1)(−0.75)=0.75 J.W=\alpha(-0.75)=(-1)(-0.75)=0.75\text{ J}.
Answer: 0.75

Topics:Work by constant and variable forcesPotential energy and conservative forces
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