a force F=-K(yi+xj), where K is a positive constant,acts on a particle moving in a x-y plane starting - Physics - Work Energy And Power

Question

a force F=-K(yi+xj), where K is a positive constant,acts on a
particle moving in a x-y plane starting from the origin yhe
particle is taken alon positive x axisto a point (a,0) and then
parallel to y axis to a point (a,a ) calculate the total work done
by force on the particle

Jyoti Pant · Accepted Answer

Solution: We know that, Work done=Force x displacement For origin to (a,0), the total work done=W =ai[-K(yi+xj)]=-Kay=0 For (a,0) to (a,a),the work done=W' =(aj+ai-ai) [-K(yi+xj)]=-Kax= $- K a^{2}$ So the total work done=W+W'=0 $- K a^{2}$ = $- K a^{2}$

a force F=-K(yi+xj), where K is a positive constant,acts on a particle moving in a x-y plane.. starting from the origin yhe particle is taken alon positive x axisto a point (a,0) and then parallel to y axis to a point (a,a.) calculate the total work done by force on the particle.