We apply Newton's second law, $\sum \vec{F} = m\vec{a}$. Let's define a coordinate system with the x-axis pointing down the incline and the y-axis perpendicular to it.

The forces acting on the mass are gravity ($mg$), the normal force ($N$) from the incline, and the kinetic friction force ($f_k$). We resolve the gravitational force into components parallel ($mg \sin\theta$) and perpendicular ($mg \cos\theta$) to the incline.

In the y-direction, there is no acceleration:

$$\sum F_y = N - mg \cos\theta = 0 \implies N = mg \cos\theta$$

This is the magnitude of the normal force exerted by the plane on the object. By Newton's third law, the force exerted by the object on the plane has the same magnitude.

In the x-direction, the net force causes acceleration $a$:

$$\sum F_x = mg \sin\theta - f_k = ma$$

The kinetic friction force is $f_k = \mu N = \mu mg \cos\theta$. Substituting this into the x-direction equation:

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

Solving for acceleration $a$:

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

To find the final speed $v$, we use kinematics. The object starts from rest ($v_0=0$) and travels a distance $d$ along the incline. From trigonometry, the distance is $d = h / \sin\theta$. Using the kinematic equation $v^2 = v_0^2 + 2ad$:

$$v^2 = 2ad = 2 \left( g(\sin\theta - \mu \cos\theta) \right) \left( \frac{h}{\sin\theta} \right)$$ $$v = \sqrt{2gh \left(1 - \mu \cot\theta \right)}$$