A particle is taken from point A to point B via the path ACB and then come back to point A via the path BDA. What is the work done by gravity on the body over this closed path, if the motion of the particle is in the vertical plane?

Here, displacement of the particle is AB, gravity is acting vertically downwards. The vertical component of AB is h upwards. Hence,W(ACB)  =-mghFor the path BDA, component of the displacement acting along vertical direction is h (downward)In this case, W(BDA)  = mgh Total work done WACB WBDA + = 0