A force F=-K (yhati + xhatj), acts on a particle moving in the x-y 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 is

A force F=-K (yhati + xhatj), acts on a particle moving in the x-y pla

1.	A force F=-K (yhati + xhatj), acts on a particle moving in the x-y 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 is
Answer» `-2Ka^2` `2Ka^2` `-Ka^2` `Ka^2` Solution :In taking the PARTICLE along the positive x-axis from (0,0) to the point (a, 0) `W_1 = int_0^a vecF .d vecx = int_0^a - K(y hati + x hatj).hati dx` `= int_0^a -K(0 hati + x hatj).hati dx = ` ZERO. And then taking the particle parallel to y-axis to the point (a,a) `W_2 = int_0^a vecF . dvecy = int_0^a - K(y hati + x hatj) .hatj dy` `= int_0^a - K(y hati + a hatj).hatj dy = -Ka int_0^a dy = -Ka^2` Total WORK done, `W = W_1 + W_2 = 0 - Ka^2 = -Ka^2`