A force f=x2y i+y2 j acts on a particle in a plane x+y=10. The work done by this force during a displacement from (0,0) to (4 m,2 m) is Joule (round off to the nearest integer)
152
W= Fx dx+ Fy dy, with y=10-x in Fx=x2y
W= 04x2(10-x) dx+ 02y2 dy
= ( 6403-64 )+ 83=152 J
Answer: 152
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