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QJEEAdv19Phc970: JEE Advanced 2019 Physics - Class 11 JEEAdv19Phc970 Solution: A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in …

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

JEE Advanced 2019 Paper 1 Offline Physics - Work Power & Energy Question 10 English

Step-by-Step Verified Solution

d = F.dr

d = α\alpha ydx + 2 α\alpha xdy

A →\to B, y = 1, dy = 0

then WA→B=∫αydx{W_{A \to B}} = \int {\alpha ydx} =α1∫01dx=α= \alpha 1\int_0^1 {dx} = \alpha

B →\to C, x = 1, dx = 0

then WB→C=2α1∫10.5dy=−2α(0.5)=−α{W_{B \to C}} = 2\alpha 1\int_1^{0.5} {dy} = - 2\alpha (0.5) = - \alpha

C →\to D, y = 0.5, dy = 0

then WC→D=∫10.5αydx=α.12∫10.5dx=−α4{W_{C \to D}} = \int_1^{0.5} {\alpha ydx} = \alpha .{1 \over 2}\int_1^{0.5} {dx} = - {\alpha \over 4}

D →\to E, x = 0.5, dx = 0

then WD→E=2α∫xdy=2α.12∫0.50dy=−α2{W_{D \to E}} = 2\alpha \int {xdy} = 2\alpha .{1 \over 2}\int\limits_{0.5}^0 {dy} = - {\alpha \over 2}

E →\to F, y = 0, dy = 0 then WEF = 0

F →\to A, x = 0, dx = 0 then WF →\to A = 0

∴\therefore W=α−α−α4−α2=−3α4W = \alpha - \alpha - {\alpha \over 4} - {\alpha \over 2} = - {{3\alpha } \over 4}

Given, α=−1\alpha = - 1

⇒W=34J=0.75J\Rightarrow W = {3 \over 4}J = 0.75J

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