A car of mass 1200 kg is moving up an inclined plane of angle 30^ with a constant speed of 10 m s ⁻¹ . A constant frictional force of 2000 N opposes the motion of the car. The power delivered by the engine of the car in kW is : (Take g = 10 m s ⁻² )

Step-by-step solution

Since the car moves with a constant speed, its acceleration is zero, which means the net force acting on it is zero. The engine must provide an upward force along the incline to balance both the component of the car's weight acting down the incline and the opposing frictional force. Component of weight down the incline = mg = 1200 10 (30^ ) = 12000 0.5 = 6000 N Total force required by the engine, F_ engine = mg + f F_ engine = 6000 + 2000 = 8000 N Power delivered by the engine, P = F_ engine v P = 8000 10 = 80000 W