A 30 kg block is on an inclined plane. Draw its free-body diagram, showing weight, normal force, and friction.

The forces acting on the block are:

• Weight ($W = mg$): Acts vertically downwards.
• Normal force ($N$): Acts perpendicular to the inclined surface, upwards.
• Friction force ($f_k$): Acts parallel to the inclined surface, opposing the motion. Since the block is moving down the incline, friction acts up the incline.

Let $\theta$ be the angle of inclination of the plane. The weight can be resolved into components parallel and perpendicular to the incline.

• Component of weight parallel to incline: $mg \sin \theta$ (down the incline).
• Component of weight perpendicular to incline: $mg \cos \theta$ (into the incline).

Here is the free-body diagram:

\begin{tikzpicture}[scale=0.8] % Inclined plane \draw (0,0) -- (6,3); \draw[dashed] (0,0) -- (6,0); \draw (5.5,0) rectangle (5.8,0.3); % Angle indicator % Block \draw[fill=lightgray] (3,1.5) rectangle (4,2.5); \node at (3.5,2) {30kg}; % Center of mass \coordinate (CM) at (3.5,2); % Weight (mg) \draw[->, thick] (CM) -- +(0,-2) node[below] {$W = mg$}; % Normal force (N) \draw[->, thick] (CM) -- ($(CM) + (-0.5,1)$) node[above left] {$N$}; % Friction force (fk) \draw[->, thick] (CM) -- ($(CM) + (-1, -0.5)$) node[above left] {$f_k$}; % Components of weight (for explanation, not usually drawn on FBD) % \draw[->, dashed] (CM) -- ($(CM) + (1, -0.5)$) node[below right] {$mg \sin\theta$}; % \draw[->, dashed] (CM) -- ($(CM) + (0.5, -1)$) node[below right] {$mg \cos\theta$}; \end{tikzpicture}

The question is cut off, but based on the context, it likely asks for the mechanical energy at a specific point. We will calculate the mechanical energy at point A.
Mechanical energy ($E$) is the sum of kinetic energy ($KE$) and potential energy ($PE$).

$$E = KE + PE$$

The formulas are:

$$KE = \frac{1}{2}mv^2$$ $$PE = mgh$$

Given:
Mass $m = 30 \text{ kg}$
Height at A, $h_A = 5 \text{ m}$
Velocity at A, $v_A = 17 \text{ m/s}$
Acceleration due to gravity $g = 9.8 \text{ m/s}^2$

Calculate kinetic energy at A:

$$KE_A = \frac{1}{2}mv_A^2 = \frac{1}{2}(30 \text{ kg})(17 \text{ m/s})^2$$ $$KE_A = \frac{1}{2}(30 \text{ kg})(289 \text{ m}^2/\text{s}^2)$$ $$KE_A = 15 \text{ kg} \times 289 \text{ m}^2/\text{s}^2 = 4335 \text{ J}$$

Calculate potential energy at A:

$$PE_A = mgh_A = (30 \text{ kg})(9.8 \text{ m/s}^2)(5 \text{ m})$$ $$PE_A = 1470 \text{ J}$$

Calculate total mechanical energy at A:

$$E_A = KE_A + PE_A = 4335 \text{ J} + 1470 \text{ J}$$ $$E_A = 5805 \text{ J}$$

The mechanical energy at point A is $5805 \text{ J}$.