A 12 kg projectile is launched vertically with an initial speed of 20 m/s. If it reaches a maximum height of 18 m, what is the work done by non-conservative forces?

The change in mechanical energy ($\Delta E_{mech}$) is the sum of the change in kinetic energy ($\Delta K$) and the change in potential energy ($\Delta U$). When non-conservative forces like air resistance are present, the work done by these forces ($W_{nc}$) is equal to the change in mechanical energy.

$$W_{nc} = \Delta E_{mech} = (K_f + U_f) - (K_i + U_i)$$

Given:
Mass of projectile, $m = 12$ kg
Initial vertical speed, $v_i = 20$ m/s
Maximum height reached, $h_f = 18$ m
Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$

At the launch point, we set the initial height $h_i = 0$.

$$K_i = \frac{1}{2}mv_i^2$$ $$K_i = \frac{1}{2}(12 \, \text{kg})(20 \, \text{m/s})^2$$ $$K_i = 6 \, \text{kg} \times 400 \, \text{m}^2/\text{s}^2$$ $$K_i = 2400 \, \text{J}$$ $$U_i = mgh_i = (12 \, \text{kg})(9.8 \, \text{m/s}^2)(0 \, \text{m})$$ $$U_i = 0 \, \text{J}$$

At the maximum height, the vertical speed of the projectile is momentarily zero, so $v_f = 0$.

$$K_f = \frac{1}{2}mv_f^2$$ $$K_f = \frac{1}{2}(12 \, \text{kg})(0 \, \text{m/s})^2$$ $$K_f = 0 \, \text{J}$$

The final height is $h_f = 18$ m.

$$U_f = mgh_f$$ $$U_f = (12 \, \text{kg})(9.8 \, \text{m/s}^2)(18 \, \text{m})$$ $$U_f = 2116.8 \, \text{J}$$

$$\Delta E_{mech} = (K_f + U_f) - (K_i + U_i)$$ $$\Delta E_{mech} = (0 \, \text{J} + 2116.8 \, \text{J}) - (2400 \, \text{J} + 0 \, \text{J})$$ $$\Delta E_{mech} = 2116.8 \, \text{J} - 2400 \, \text{J}$$ $$\Delta E_{mech} = -283.2 \, \text{J}$$

$$\Delta E_{mech} = -283.2 \, \text{J} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = -0.2832 \, \text{kJ}$$

Rounding to two decimal places, this is $-0.28$ kJ.

The final answer is $\text{a. -0.28 kJ}$.