Question of Physics

Question

Force F acting on a particle is given by

F= -kyi-kxj

where k is a constant. Find work done by F when the particle is moving from the origin A (0,0) to point B(a,b)

1)along path 1 and

2) along path 2.

Path 1= y=b(1-e^-kx)/(1-e^-ka)

Path 2= y= bx³/a³

Answer 1

Work = FdotdS .... dS=dxi+dyj ... dy = derivative of path*dx ... Integrate FdotdS from 0 to a to find Work = -kab(1-e^-ka-ka/2)/(1-e^-ka)