Interpret the figure and motion

The particle starts from point $A$ and, after a slight displacement to the right, moves to point $B$.

Since the question is from work, power and energy, the natural method is to use conservation of mechanical energy:

$\text{Loss in gravitational potential energy} = \text{Gain in kinetic energy}$

Apply energy conservation

Let the vertical drop from $A$ to $B$ be $h$.

Then, $mgh = \frac{1}{2}mv^2$

Mass cancels out: $gh = \frac{v^2}{2}$

so, $v = \sqrt{2gh}$

Use the height difference from the figure

From the figure, the particle drops through a vertical height of $h = 5\,\text{m}$

With $g = 10\,\text{m/s}^2$, $v = \sqrt{2 \cdot 10 \cdot 5} = \sqrt{100} = 10\,\text{m/s}$

Final answer

$x = 10$

Comparison with stored answer

Stored correct answer = $10$.

Our derived answer also equals $10$, so they agree.