To solve this problem we will apply the work theorem which is expressed as the force applied to displace a body. Considering that body strength is equivalent to weight, we will make the following considerations
[tex]\text{Mass of the child} = m = 25kg[/tex]
[tex]\text{Acceleration due to gravity} = g = 9.81m/s^2[/tex]
[tex]\text{Height lifted} = h = 0.80m (Upward)[/tex]
Work done to upward the object
[tex]W = mgh[/tex]
[tex]W = (25)(9.81)(0.8)[/tex]
[tex]W = 196.2J[/tex]
Horizontal Force applied while carrying 10m,
[tex]F = 0N[/tex]
[tex]W = 0J[/tex]
Height descended in setting the child down
[tex]h' = -0.8m (Downwoard)[/tex]
[tex]W = mgh'[/tex]
[tex]W = (25)(9.81)(-0.80)[/tex]
[tex]W = -196.2J[/tex]
For full time, assuming that the total value of work is always expressed in terms of its symbol, it would be zero, since at first it performs the same work that is later complemented in a negative way.
