The work done by the person, $W_p$, equals the work done by the tension force, which changes the mechanical energy of the cargo. Let $h = 2.0$ m.

[Q1] For constant velocity, the pulling force $F$ equals the gravitational force $mg$. The work done is:

$$W_p = Fh = mgh$$ $$W_p = (60 \text{ kg})(9.8 \text{ m/s}^2)(2.0 \text{ m})$$

[Q2] By the work-energy theorem, the net work done by all forces equals the change in kinetic energy: $W_{net} = \Delta K$. The net work is the sum of work done by the person ($W_p$) and work by gravity ($W_g = -mgh$).

$$W_p + W_g = K_f - K_i$$ $$W_p - mgh = \frac{1}{2}mv_f^2 - 0$$

Solving for the work done by the person:

$$W_p = mgh + \frac{1}{2}mv_f^2$$ $$W_p = (60 \text{ kg})(9.8 \text{ m/s}^2)(2.0 \text{ m}) + \frac{1}{2}(60 \text{ kg})(20 \text{ m/s})^2$$