In this scenario, we are analyzing the motion of a 20-kilogram block being pushed up an incline with a force of 110 Newtons. To determine the acceleration of the block, we first need to establish a free body diagram that includes the gravitational force acting downwards, the normal force acting perpendicular to the surface, and the applied force pushing the block up the ramp. Additionally, we must account for friction, which opposes the direction of motion.

The gravitational force can be broken down into two components: the component acting parallel to the incline, denoted as \( mg_x \), and the component acting perpendicular to the incline, denoted as \( mg_y \). The parallel component is calculated using the equation \( mg_x = mg \sin(\theta) \), where \( m \) is the mass of the block, \( g \) is the acceleration due to gravity (approximately 9.8 m/s²), and \( \theta \) is the angle of the incline. For this problem, with \( m = 20 \) kg and \( \theta = 15^\circ \), we find \( mg_x = 20 \times 9.8 \times \sin(15^\circ) \approx 50.7 \) N.

Next, we compare the applied force (110 N) with the gravitational force component (50.7 N) to determine the direction of friction. Since the applied force is greater, friction will act down the ramp. To find the type of friction, we calculate the maximum static friction force using the equation \( f_s^{max} = \mu_s \cdot N \), where \( \mu_s \) is the coefficient of static friction (0.3) and \( N \) is the normal force, calculated as \( N = mg_y = mg \cos(\theta) \). Thus, \( N = 20 \times 9.8 \times \cos(15^\circ) \approx 56.8 \) N, leading to \( f_s^{max} = 0.3 \times 56.8 \approx 17.04 \) N.

Since the net force acting on the block (110 N - 50.7 N = 59.3 N) exceeds the maximum static friction, the block experiences kinetic friction instead. The kinetic friction force is calculated using \( f_k = \mu_k \cdot N \), where \( \mu_k \) is the coefficient of kinetic friction (0.2). Therefore, \( f_k = 0.2 \times 56.8 \approx 11.36 \) N.

Now, we can apply Newton's second law, \( F = ma \), along the direction of motion. The equation becomes \( F - mg_x - f_k = ma \). Substituting the known values gives us:

110 N - 50.7 N - 11.36 N = 20 kg \cdot a.

This simplifies to:

48.94 N = 20 kg \cdot a.

Solving for acceleration \( a \) yields:

a = \frac{48.94}{20} \approx 2.447 \, \text{m/s}^2.

Thus, the block accelerates up the ramp at approximately 2.45 m/s², indicating that the applied force is sufficient to overcome both the gravitational pull and the kinetic friction acting against it.