Step 1: Identify the forces acting on the box and resolve the applied force into components. The forces acting on the box are: • Weight ($W$) acting downwards. $W = 400$ N. • Applied force ($P$) acting at an angle of $30.0^\circ$ below the horizontal. $P = 100$ N. • Horizontal component of $P$: $P_x = P \cos \theta$ • Vertical component of $P$: $P_y = P \sin \theta$ (acting downwards) • Normal force ($N$) acting upwards from the floor. • Kinetic friction force ($f_k$) acting horizontally, opposite to the direction of motion. Since the box is moving at a constant speed, the net force in both the horizontal and vertical directions is zero. Step 2: Apply Newton's second law in the vertical direction to find the normal force ($N$). The forces in the vertical direction are the normal force ($N$) upwards, the weight ($W$) downwards, and the vertical component of the applied force ($P_y$) downwards. $$ \sum F_y = 0 $$ $$ N - W - P_y = 0 $$ $$ N = W + P_y $$ Substitute $P_y = P \sin \theta$: $$ N = W + P \sin \theta $$ Substitute the given values: $W = 400$ N, $P = 100$ N, $\theta = 30.0^\circ$. $$ N = 400 \text{ N} + (100 \text{ N}) \sin(30.0^\circ) $$ $$ N = 400 \text{ N} + (100 \text{ N}) (0.5) $$ $$ N = 400 \text{ N} + 50 \text{ N} $$ $$ N = 450 \text{ N} $$ Step 3: Apply Newton's second law in the horizontal direction to find the kinetic friction force ($f_k$). The forces in the horizontal direction are the horizontal component of the applied force ($P_x$) in the direction of motion and the kinetic friction force ($f_k$) opposite to the direction of motion. $$ \sum F_x = 0 $$ $$ P_x - f_k = 0 $$ $$ f_k = P_x $$ Substitute $P_x = P \cos \theta$: $$ f_k = P \cos \theta $$ Substitute the given values: $P = 100$ N, $\theta = 30.0^\circ$. $$ f_k = (100 \text{ N}) \cos(30.0^\circ) $$ $$ f_k = (100 \text{ N}) (0.8660) $$ $$ f_k = 86.60 \text{ N} $$ Step 4: Calculate the coefficient of kinetic friction ($\mu_k$). The kinetic friction force is related to the normal force by the formula: $$ f_k = \mu_k N $$ Rearrange to solve for $\mu_k$: $$ \mu_k = \frac{f_k}{N} $$ Substitute the calculated values for $f_k$ and $N$: $$ \mu_k = \frac{86.60 \text{ N}}{450 \text{ N}} $$ $$ \mu_k \approx 0.1924 $$ Comparing this value to the given options: A. 0.19 B. 0.24 C. 0.40 D. 0.22 The calculated value $0.1924$ is closest to $0.19$. The final answer is $\boxed{\text{A. } 0.19}$.