The forces acting on the block are the applied force $F$, gravity $F_g = mg$, the normal force $N$, and the kinetic friction force $f_k$. Since the velocity is constant, the net force is zero.

Applying Newton's second law in the vertical and horizontal directions:

$$ \Sigma F_y = N + F \sin\theta - mg = 0 \implies N = mg - F \sin\theta $$ $$ \Sigma F_x = F \cos\theta - f_k = 0 \implies f_k = F \cos\theta $$

[Q1] From the free-body diagram, there are four forces acting on the block.

[Q2] The work done by a constant force is $W = \vec{F} \cdot \vec{d} = Fd \cos\phi$, where $\phi$ is the angle between the force and displacement vectors. Work done by the applied force $F$:

$$ W_F = F d \cos\theta $$

Work done by the friction force $f_k$ (acts opposite to displacement, $\phi=180^\circ$):

$$ W_{f_k} = -f_k d = -(F \cos\theta) d = -F d \cos\theta $$

The normal force and gravity are perpendicular to the displacement, so they do no work:

$$ W_N = 0 $$ $$ W_g = 0 $$

The net work is the sum of the work done by all forces. Since the net force is zero (constant velocity), the net work is also zero.

$$ W_{net} = W_F + W_{f_k} + W_N + W_g = F d \cos\theta - F d \cos\theta + 0 + 0 = 0 $$

[Q3] The kinetic friction force is defined as $f_k = \mu_k N$. Substituting the expressions for $f_k$ and $N$ from Newton's laws:

$$ F \cos\theta = \mu_k (mg - F \sin\theta) $$

Solving for the coefficient of kinetic friction $\mu_k$:

$$ \mu_k = \frac{F \cos\theta}{mg - F \sin\theta} $$