Q.

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=-1 N m-1. The work done on the particle by this force F will be _____ Joule.                  [2019]


Ans.

(0.75)

Given : Force, F=(αyi^+2αxj^)

and α=-1 N m-1

We know that dW=F·dr=(αyi^+2αxj^)·(dxi^+dyj^)

 dW=αydx+2αxdy

Work done

From AB, dy=0, as y=1

 W1=01αydx=α01dx=α

From BC, dx=0, as x=1

 W2=10.52αxdy=10.52αdy=2α(0.5-1)=-α

From CD, dy=0, as y=0.5

 W3=10.5α×0.5dx=α2(0.5-1)=-α4

From DE, dx=0, as x=0.5

 W4=0.502α×0.5dy=α(0-0.5)=-α2

From EF, dy=0, as y=0

 W5=α×0dx=0

From FA, dx=0, as x=0

 W6=2α×0dy=0

 Total work done W=W1+W2+W3+W4+W5+W6

W=α-α-α4-α2=-3α4=-3(-1)4=0.75J