5.1 State the principle of conservation of mechanical energy in words.

Initial state (P): $v_P = 0$ m/s, $h_P = 0$ m.
Final state (R): $v_R = 0$ m/s, $h_R = 40$ cm $= 0.4$ m.
Forces doing work from P to R: applied force $F$, frictional force $f_k$, and gravitational force $F_g$.

The Work-Energy Theorem states that the net work done on an object equals its change in kinetic energy: $W_{net} = \Delta K$.
Since the object starts from rest at P and stops at R, $\Delta K = K_R - K_P = 0 - 0 = 0$.
Therefore, $W_{net} = 0$.

The net work is the sum of the work done by all forces:

Note: The applied force $F$ and frictional force $f_k$ only act from P to Q. The gravitational force $F_g$ does work over the vertical displacement from Q to R.

Given:
$F = 80$ N
$\Delta x = 4$ m (distance from P to Q)
$f_k = 45$ N
$m = 20$ kg
$g = 9.8$ m/s$^2$
$h_R = 0.4$ m

The magnitude of the angle $\theta$ is $\boxed{\text{36.13}^\circ}$.

5.3.2 Power delivered by force F if the time taken for the object to move from point P to point Q is 2.86 seconds.

Using the value of $\theta$ from 5.3.1:

Power $P$ is defined as the rate at which work is done:

Given $t = 2.86$ s:

The power delivered by force F is $\boxed{\text{90.35 W}}$.