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 \( a = 3g \), and the angle of inclination is \( 30^\circ \). <br /><br />### Step-by-step solution:<br /><br />1. **Acceleration components**:<br /> - The acceleration of the plane (\( a = 3g \)) has two components relative to the inclined plane:<br /> - Parallel to the incline: \( a_\parallel = 3g \sin 30^\circ = 3g \cdot \frac{1}{2} = 1.5g \).<br /> - Perpendicular to the incline: \( a_\perp = 3g \cos 30^\circ = 3g \cdot \frac{\sqrt{3}}{2} = 1.5\sqrt{3}g \).<br /><br />2. **Forces on the block**:<br /> - The block experiences gravitational force \( mg \) vertically downward.<br /> - Relative to the inclined plane, the effective acceleration of the block is determined by combining the effects of gravity and the plane's acceleration.<br /><br />3. **Relative acceleration**:<br /> - The net acceleration of the block relative to the plane is given by:<br /> \[<br /> a_{\text{relative}} = \sqrt{(g \sin 30^\circ + a_\parallel)^2 + (g \cos 30^\circ - a_\perp)^2}.<br /> \]<br /> - Substituting values:<br /> - \( g \sin 30^\circ = g \cdot \frac{1}{2} = 0.5g \),<br /> - \( g \cos 30^\circ = g \cdot \frac{\sqrt{3}}{2} = 0.866g \),<br /> - \( a_\parallel = 1.5g \),<br /> - \( a_\perp = 1.5\sqrt{3}g = 2.598g \).<br /><br /> - Plugging these into the formula:<br /> \[<br /> a_{\text{relative}} = \sqrt{(0.5g + 1.5g)^2 + (0.866g - 2.598g)^2}.<br /> \]<br /> Simplify:<br /> \[<br /> a_{\text{relative}} = \sqrt{(2g)^2 + (-1.732g)^2}.<br /> \]<br /> \[<br /> a_{\text{relative}} = \sqrt{4g^2 + 3g^2}.<br /> \]<br /> \[<br /> a_{\text{relative}} = \sqrt{7g^2}.<br /> \]<br /> \[<br /> a_{\text{relative}} = \sqrt{7}g \approx 2.645g.<br /> \]<br /><br />4. **Convert to numerical value**:<br /> - Since \( g = 9.8 \, \text{m/s}^2 \), the relative acceleration is approximately:<br /> \[<br /> a_{\text{relative}} = 2.645 \cdot 9.8 \approx 25.91 \, \text{m/s}^2.<br /> \]<br /><br />Thus, the closest option is **(c) \( 2.80g \)**.