Answer:
Work done = 0 J
Step-by-step explanation:
work done= ∫ F. dr
= [tex]\int\limits^2_0 {x} \, dx[/tex] + [tex]\int\limits^2_2 {x} \, dx[/tex] + [tex]\int\limits^0_2 {x} \, dx[/tex] + [tex]\int\limits^0_0 {x} \, dx[/tex] + [tex]\int\limits^0_0 {y} \, dy[/tex] + [tex]\int\limits^5_0 {y} \, dy[/tex] + [tex]\int\limits^5_5 {y} \, dy[/tex] + [tex]\int\limits^0_5 {y} \, dy[/tex] + [tex]\int\limits^0_0 {z} \, dz[/tex] + [tex]\int\limits^1_0 {z} \, dz[/tex] + [tex]\int\limits^1_1 {z} \, dz[/tex] + [tex]\int\limits^0_1 {z} \, dz[/tex]
Work done= x²/2 + y²/2 + z²/2
Applying integral limits for entire pathway
Work done= 2 + 0 -2 + 0 + 0+ 25/2 - 25/2 + 0 + 1/2 + 0 - 1/2
Work done = 0 J
