The work done by a constant force $\vec{F}$ during a displacement $\vec{d}$ is given by the dot product $W = \vec{F} \cdot \vec{d} = |\vec{F}||\vec{d}|\cos\phi$, where $\phi$ is the angle between the force and displacement vectors.

[Q1] The displacement vector $\vec{d}$ points up the ramp.

1. Work by applied force $\vec{F_a}$: The force is horizontal, and the displacement is at an angle $\theta$ to the horizontal. The angle between them is $\theta$. $$W_a = F_a d \cos\theta$$
2. Work by gravitational force $\vec{F_g}$: The force $m\vec{g}$ is vertically downward. The vertical component of the displacement is $d\sin\theta$ upward. The work done is negative. $$W_g = -mg(d\sin\theta) = -mgd\sin\theta$$
3. Work by normal force $\vec{F_N}$: The normal force is perpendicular to the ramp surface, while the displacement is parallel to it. The angle between them is $90^\circ$. $$W_N = F_N d \cos(90^\circ) = 0$$
4. Net work $W_{net}$: The net work is the sum of the work done by all individual forces. $$W_{net} = W_a + W_g + W_N = F_a d \cos\theta - mgd\sin\theta$$

[Q2] According to the work-energy theorem, the net work done on an object equals the change in its kinetic energy: $W_{net} = \Delta K = K_f - K_i$. The book starts from rest, so its initial kinetic energy $K_i=0$. The final kinetic energy is $K_f = \frac{1}{2}mv_f^2$.

Substituting the expression for $W_{net}$ from Q1 and solving for $v_f$: