P01R2015 – Acceleration on an incline

We draw the acceleration vector $\overrightarrow{g}$ acting on the block and show its $x$ and $y$ components in the figure below.

From trigonometry, we derive that these components can be expressed as follows

\begin{aligned}
{\mathrm{cos} \left(\theta \right)\ }&=\frac{g_y}{g}\ \ \ \ \ \Rightarrow \ \ \ \ \ g_y=g{\mathrm{cos} \left(\theta \right)\ } \\
\\
{\mathrm{sin} \left(\theta \right)\ }&=\frac{g_x}{g}\ \ \ \ \ \Rightarrow \ \ \ \ \ g_x=g{\mathrm{sin} \left(\theta \right)\ }
\end{aligned}

Thus, the $x$-component of the acceleration due to gravity $\overrightarrow{g}$ is equal to

\boxed{a_x=gsin\left(\theta \right)}

The block slides down from rest with an acceleration $a_x=gsin\left(\theta \right)$ in the $x$ direction, traveling a distance $D$. Its speed at the bottom of the incline is therefore given by

\begin{aligned}
v^2_{fx}=v^2_{0x}+2a_x\mathrm{\Delta }x\ \ \ \ \ &\Rightarrow \ \ \ \ v^2_{fx}=2g{\mathrm{sin} \left(\theta \right)\ }D \\
\\
&\Rightarrow \ \ \ \ \ v_{fx}=\sqrt{2g{\mathrm{sin} \left(\theta \right)\ }D}
\end{aligned}

Thus, the speed of the block at the bottom of the incline is equal to

\boxed{v_{fx}=\sqrt{2g{\mathrm{sin} \left(\theta \right)\ }D}=\sqrt{2\cdot 9.8\cdot {\mathrm{sin} \left(30\right)\ }\cdot 20}=14\ m/s}

The acceleration that the block experiences is constant throughout its entire motion and equal to the acceleration due to gravity $\overrightarrow{g}$ which is always directed vertically downward (toward the center of the Earth).

The $x$-component of $\overrightarrow{g}$ always points toward the base of the incline and therefore explains why the block slows down as it slides up the incline and then speeds up as it slides back down.