11position of the particle is ai + aj
W= F. S
W = -kay-kax
W= -ka(y + x)
Is it rite?
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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4 Answers
11
33Work done from (0,0) to (a,0)
W=0∫aF.dx
=-k∫(yi+xj).dxi
=-k∫ydx
=-ky∫dx=0.. (y=0)
From (a,0) to (a,a)
-∫kF.dy
=-kx0∫ady
=-ka[y]0a
=-ka2
33Same question but different approach here:
http://targetiit.com/iit_jee_forum/posts/conceptual_physics_3695.html
24@ virang......apply W= F. S. if and only if force is constant,,,,here we can clearly see that force is variable so we apply ....dw=∫f.ds
