Use the Work-Energy Theorem to solve the following: A 25 kg crate is pulled 3.5 m up a 30 degree incline. If the crate starts at 12 m/s and ends at 10.8 m/s, what is the magnitude of the force pulling

Mass of the crate, $m = 25 \text{ kg}$
Initial speed at point Q, $v_Q = 12 \text{ m/s}$
Final speed at point R, $v_R = 10.8 \text{ m/s}$
Distance moved along the incline, $d = 3.5 \text{ m}$
Angle of inclination, $\alpha = 30^\circ$
Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

The vertical height gained, $h = d \sin(\alpha)$.

The work done by the applied force $F$ ($W_F$) is the non-conservative work.

The work done by force $F$ is also given by $W_F = Fd \cos\theta$, where $\theta$ is the angle between the force and displacement. Since $F$ is parallel to the incline and the displacement is up the incline, $\theta = 0^\circ$, so $\cos(0^\circ) = 1$.

Rounding to three significant figures:

The magnitude of force $F$ is $\boxed{\text{24.8 N}}$.