a forceF\vec{} = -k(yi\dot{} + xj\dot{}) (where K is a positive constant acts on a particle moving in the xy plane starting from the origin the particle is taken along the positive x axis to the point (a,0) and then parallel to the y axis to the point (a,a) . the total work done by the force on the particle is
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2 Answers
33path from 0,0 to a,a
dr=dxi+dyj=dxi+dyj
F=-k(yi+xj)
dW=F.dr
=-k(yi+xj).(dxi+dyj)
=-k(ydx+xdy)
=-∫kd(xy) ..here it shows that force is conservative... from point P(x0,y0) to again back there work done is zero...
so directly integrating from initial to final point....
=-k0∫a2d(xy)
=-kxy]0a2
=-ka2
33Same question, different approach here:
http://targetiit.com/iit_jee_forum/posts/conceptual_physics_3696.html
