A 2.2-kg block rests on a 30.0° incline as shown in the figure.The coefficient of static friction between the block and the incline is 0.63. There is a horizontal force force F  of 6.4 N applied to the block. This is the maximum force possible without causing the block to slip. What is the normal force exerted on the block by the inclined surface?Express your answer in N, to at least one digit after the decimal point.

To solve this problem, we need to consider the forces acting on the block.

1. First, we calculate the weight of the block, which is the mass times the acceleration due to gravity. In this case, it's 2.2 kg * 9.8 m/s^2 = 21.56 N. This force acts vertically downward.

2. The weight of the block can be resolved into two components: one perpendicular to the incline (the normal force) and one parallel to the incline. The component of the weight perpendicular to the incline is given by W*cos(θ), where θ is the angle of the incline. So, the normal force is 21.56 N * cos(30.0°) = 18.67 N.

3. The component of the weight parallel to the incline is given by W*sin(θ). This is 21.56 N * sin(30.0°) = 10.78 N. This force is trying to make the block slide down the incline.

4. The frictional force acting on the block is given by the coefficient of static friction times the normal force, which is 0.63 * 18.67 N = 11.78 N. This force is opposing the component of the weight trying to make the block slide down the incline.

5. The horizontal force applied to the block is 6.4 N. This force can be resolved into two components: one parallel to the incline and one perpendicular to the incline. The component of the force parallel to the incline is 6.4 N * sin(30.0°) = 3.2 N. This force is trying to make the block slide up the incline.

6. The component of the force perpendicular to the incline is 6.4 N * cos(30.0°) = 5.54 N. This force is adding to the normal force.

7. The total normal force is the sum of the component of the weight perpendicular to the incline and the component of the force perpendicular to the incline, which is 18.67 N + 5.54 N = 24.21 N.

So, the normal force exerted on the block by the inclined surface is 24.21 N.