In a children's park, an inclined plane is built with an angle of...

The forces acting on the boy along the inclined plane are the component of gravity downwards and the frictional force upwards.

The equation of motion for the boy sliding down the incline is given by:

$ma = mg \sin \theta - f_k$

where $f_k = \mu N$ is the force of kinetic friction and $N = mg \cos \theta$ is the normal reaction.

Substituting $f_k$ into the equation, we get:

$ma = mg \sin \theta - \mu mg \cos \theta$

$a = g(\sin \theta - \mu \cos \theta)$

Given values are $g = 10 \text{ m/s}^2$, $\theta = 45^{{}^{{}^{\circ}}}$, and $\mu = 0.6$.

Substituting these values:

$a = 10$\left( \sin45^{{}^{{}^{\circ}}} - 0.6 \cos 45^{{}^{{}^{\circ}}} \right)$<br><br>$a = 10 \left( \dfrac{1}{\sqrt{2}} -$0.6$\times \dfrac{1}{\sqrt{2}} \right)

$a =$\dfrac{10}{\sqrt{2}} ($1 - 0.6)$

$a =$\dfrac{10}{\sqrt{2}} \times$0.4$

$a =$\dfrac{4}{\sqrt{2}}

$a =$2\sqrt{2} \text{ m/s}^2