10.A block of mass m is placed at the highest point of a triangular plane moving on a horizontal surface with an acceleration of a=3g as shown in the figure . Assume all surfaces are smooth and the plane makes a 30^circ angle with the floor . What is the acceleration (inm/s^2) of the block relative to the plane? (Attention:an incorrect answer will be penalized 2 pts.) (a.) 2.208 b) 3.11g c) 2.80g d) 0.63 g e) 1.40 g

To solve this problem, we analyze the forces acting on the block relative to the inclined plane. The triangular plane is accelerating horizontally with , and the angle of inclination is . ### Step-by-step solution:1. **Acceleration components**: - The acceleration of the plane ( ) has two components relative to the inclined plane: - Parallel to the incline: . - Perpendicular to the incline: .2. **Forces on the block**: - The block experiences gravitational force vertically downward. - Relative to the inclined plane, the effective acceleration of the block is determined by combining the effects of gravity and the plane's acceleration.3. **Relative acceleration**: - The net acceleration of the block relative to the plane is given by: - Substituting values: - , - , - , - . - Plugging these into the formula: Simplify: 4. **Convert to numerical value**: - Since , the relative acceleration is approximately: Thus, the closest option is **(c) **.