First slide
Work done by Diff type of forces

A force  F¯=(y2x2+z2)i^+(3xy5z)j^+4zk^ is applied on a particle from the point (0,0,0) to the point (2,4,0) in the path shown . If  WA  is the work done by the force in the paths A, then WA =? 


The work done by the external force is F.dS . Now here in this case the z coordinate of the motion of the particle is zero. F=(y2-x2)i^+3xyj^

Hence WA=02x2dx+043xy dy =[x33  ]02+[3y2]04=83+48=1363

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