This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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here are the solutions to your physics questions: 5.1 If the car moves up the inclined plane at CONSTANT velocity, its kinetic energy does not change. According to the Work-Energy Theorem, the net work done on an object is equal to the change in its kinetic energy. Therefore, the net work done on the car is zero. 5.2 Weight (F_g or mg): Acts vertically downwards from the center of mass. Normal Force (F_N or N): Acts perpendicular to the inclined surface, upwards from the surface. Engine Force (F_engine): Acts parallel to the inclined surface, upwards along the incline (in the direction of motion). Kinetic Friction Force (f_k): Acts parallel to the inclined surface, downwards along the incline (opposing motion). `mermaid graph TD A[Car] -->|F_engine (up incline)| B(Motion) A -->|F_N (perpendicular to incline)| C(Surface) A -->|mg (vertically down)| D(Earth) A -->|f_k (down incline)| E(Surface) style A fill:#fff,stroke:#333,stroke-width:2px style B fill:#fff,stroke:#333,stroke-width:0px style C fill:#fff,stroke:#333,stroke-width:0px style D fill:#fff,stroke:#333,stroke-width:0px style E fill:#fff,stroke:#333,stroke-width:0px subgraph Free Body Diagram direction TD Car_CM(Center of Mass) Car_CM -->|F_engine| Up_Incline Car_CM -->|F_N| Perpendicular_Up Car_CM -->|mg| Vertically_Down Car_CM -->|f_k| Down_Incline end ` (Note: The mermaid diagram is a textual representation. In a hand-drawn diagram, these would be arrows originating from the center of mass of the car.) 5.3.1 The Work-Energy Theorem states that the net work done on an object is equal to the change in its kinetic energy. 5.3.2 Step 1: Calculate the normal force (N). The normal force is perpendicular to the incline and balances the perpendicular component of gravity. N = mg Given: m = 700 kg, g = 9.8 m/s^2, = 30^. N = (700 kg)(9.8 m/s^2) 30^ N = 6860 N × 0.8660 N = 5940.76 N Step 2: Calculate the kinetic friction force (f_k). f_k = _k N Given: _k = 0.32. f_k = 0.32 × 5940.76 N f_k = 1901.04 N Step 3: Calculate the work done by each force. The car moves a distance d = 70 m up the incline. Work done by the engine (W_engine): W_engine = F_engine · d Given: F_engine = 6000 N. W_engine = (6000 N)(70 m) W_engine = 420000 J Work done by gravity (W_g): Gravity acts downwards, opposing the upward motion. The component of gravity parallel to the incline is mg . W_g = -mg · d W_g = -(700 kg)(9.8 m/s^2) 30^ (70 m) W_g = -6860 N × 0.5 × 70 m W_g = -240100 J Work done by friction (W_f): Friction opposes motion, so it acts down the incline. W_f = -f_k · d W_f = -(1901.04 N)(70 m) W_f = -133072.8 J Step 4: Calculate the net work done (W_net). W_net = W_engine + W_g + W_f W_net = 420000 J - 240100 J - 133072.8 J W_net = 46827.2 J Step 5: Apply the Work-Energy Theorem to find the final velocity (v_f). W_net = K = (1)/(2)mv_f^2 - (1)/(2)mv_i^2 The car starts from rest, so v_i = 0. W_net = (1)/(2)mv_f^2 46827.2 J = (1)/(2)(700 kg)v_f^2 46827.2 J = 350 kg · v_f^2 v_f^2 = 46827.2 J350 kg v_f^2 = 133.792 v_f = sqrt(133.792) v_f = 11.57 m/s What's next?