Mobile inclined plane

When acceleration gets confusing, I go back to position and work my way up. Let the position of the top of the plane, relative to the inertial frame, be $overrightarrow r_p(t)$. And let the block's position relative to the top of the plane be $overrightarrow r_{b|p}(t)$. Then, the inertial position $overrightarrow r_b(t)$ of the block is $overrightarrow r_{b|p}(t) + overrightarrow r_p(t)$. It's acceleration is the second derivative,

$overrightarrow a_b(t) = overrightarrow r_{b|p}''(t) + overrightarrow r_p''(t) = overrightarrow a_{b|p} + overrightarrow a_p$

So Newton's 2nd law, assuming only normal force and gravity, says

$overrightarrow F_N + moverrightarrow g = m(overrightarrow a_{b|p} + overrightarrow a_p)$

Now, we know the directions of all four vectors: $overrightarrow F_N$ is perpendicular to the plane, $moverrightarrow g$ is vertical, $overrightarrow a_{b|p}$ is along the plane, and $overrightarrow a_p$ is horizontal. So you can pick your x and y axes, break the equation into components, and solve. But you can see that it wouldn't work to solve it in the plane's frame of reference, because the forces themselves are responsible for supplying that extra horizontal acceleration. In fact, since I assumed no friction, the only force with a horizontal component is the normal force, so the more the plane accelerates, the higher the normal force. And that means the vertical component of the normal force increases as well, until it actually outweighs gravity, and the block moves up the ramp instead of down - which you can probably imagine; it's like you're sliding the ramp right out from under the block. But you would never get that result by solving in the plane frame.

Basically this is an example of the general principle that Newton's laws are only valid in an inertial frame!