We are asked to find the block's speed when it reaches the bottom of the incline. Let's break this down step-by-step.

Given:

• Mass of the block, $m = 2.0 \, \text{kg}$
• Initial speed, $v_i = 1.0 \, \text{m/s}$
• Height above the bottom, $h = 3.0 \, \text{m}$
• Length of the incline, $d = 5.0 \, \text{m}$
• Coefficient of kinetic friction, $\mu_k = 0.75$
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$

Steps:

1. Calculate the gravitational force component along the incline: The block is sliding down the incline due to gravity, which has a component along the incline.

   The angle of the incline $\theta$ can be found using:

   \quad \Rightarrow \quad \theta = \sin^{-1}(0.6)$$ Now, the component of gravitational force along the incline: $$F_{\text{gravity}} = mg \sin \theta$$

2. Frictional force: The frictional force opposing the motion is: $F_{\text{friction}} = \mu_k mg \cos \theta$ Here, $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - 0.6^2} = 0.8$.

3. Net force and acceleration: The net force along the incline is the difference between the gravitational force and the frictional force: $F_{\text{net}} = F_{\text{gravity}} - F_{\text{friction}} = mg \sin \theta - \mu_k mg \cos \theta$ The net acceleration $a$ is: $a = \frac{F_{\text{net}}}{m} = g (\sin \theta - \mu_k \cos \theta)$

4. Using kinematic equations: To find the final velocity $v_f$, we use the kinematic equation: $v_f^2 = v_i^2 + 2ad$ Solving for $v_f$: $v_f = \sqrt{v_i^2 + 2ad}$

Now, let’s calculate the individual components.

1. Calculate the angle $\theta$:

$\theta = \sin^{-1}(0.6) = 36.87^\circ$

2. Gravitational and frictional forces:

$F_{\text{gravity}} = 2.0 \times 9.8 \times \sin(36.87^\circ) = 11.76 \, \text{N}$ $F_{\text{friction}} = 0.75 \times 2.0 \times 9.8 \times \cos(36.87^\circ) = 11.76 \, \text{N}$

3. Net force and acceleration:

$F_{\text{net}} = 11.76 - 11.76 = 0 \, \text{N}$ [ a = g (\sin \theta - \mu_k \cos \theta) =