(I) A 75.0-kg firefighter climbs a flight of stairs 28.0 m high. How much work does he do?

In this problem, the amount of force needed to lift the firefighter is equivalent to its weight.

The net work done will be equal to the change in gravitational potential energy.

The relation of work done (W) is given by:

\(\begin{aligned}W &= mg(h - 0)\\W &= mgh\end{aligned}\)

Here, g is the gravitational acceleration, and h is the height of the stairs.

Given data:

The mass of the firefighter is\(m = 75\;{\rm{kg}}\).

The height of the stairs is \(h = 28\;{\rm{m}}\).

On plugging the values in the above relation, you get:

\(\begin{aligned}W &= \left( {75\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {28\;{\rm{m}}} \right)\\W &= 2.06 \times {10^4}\;{\rm{J}}\end{aligned}\)

Thus, \(W = 2.06 \times {10^4}\;{\rm{J}}\) is the correct answer.