Question

You lift a 25-kg child 0.80 m, slowly carry him 10 m to the playroom, and finally set him back down 0.80 m onto the playroom floor. What work do you do on the child for each part of the trip and for the whole trip

Answer 1

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

$\text{Mass of the child} = m = 25kg$

$\text{Acceleration due to gravity} = g = 9.81m/s^2$

$\text{Height lifted} = h = 0.80m (Upward)$

Work done to upward the object

$W = mgh$

$W = (25)(9.81)(0.8)$

$W = 196.2J$

Horizontal Force applied while carrying 10m,

$F = 0N$

$W = 0J$

Height descended in setting the child down

$h' = -0.8m (Downwoard)$

$W = mgh'$

$W = (25)(9.81)(-0.80)$

$W = -196.2J$

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.