A particle is moved along a path A B - B C - C D - D E - E F - F A , as shown in figure, in presence of a…

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^N, where x and y are in meter and α=-1N/m-1. The work done on the particle by this force F will be ____ Joule.

Solution

dw=Fdr
dw=aydx+2axdy
Now, total work done in whole path is given by,
W=WAB+WBC+WCD+WDE+WEF+WFA
AB,y=1,dy=0,WAB=αydx=α101dx=α
BC,x=1,dx=0,WBC=2α110.5dy=-2α0.5=-α
CD,y=0.5,dy=0,WCD=10.5αydx=α1210.5dx=-α4
DE,x=0.5,dx=0,WDE=2αxdy=2α1210.5dy=-α2
EF,y=0,dy=0,WEF=0
FA,x=0,dx=0,WEA=0
W=α-α-α4-α2=-3α4
Given α=-1W=+34J=0.75J !

Asked in: JEE Advanced 2019 (Paper 1)

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