The front of a lorry can be considered a flat vertical surface of area 8.0m2. The average pressure due to air resistance is 150 Pa when the lorry travels at a constant speed of 16 m s-1. You may assume the air resistance force acts perpendicular to the front of the lorry. When the lorry travels at 16 m s-1. a) calculate the force due to the air resistance force due to air resistance = ................ N (2 Marks) b) calculate the power dissipated in overcoming the air resistance. power = ..................... W (3 Marks) c) State and explain how the values of the quantities calculated above would change if the lorry increased its speed. (3 Marks) (Marks available: 8) An electric motor drives a pulley which lifts a mass of 6kg at a constant speed of 3 m s-1 (g = 9.81 m s-2) a) What force is needed to lift the mass? (1 Mark) b) What is the power output of the motor? (2 Marks) c) If the motor is 60% efficient, what is the power input from the electricty supply? (2 Marks) (Marks available: 5) Answer Answer outline and marking scheme for question: a) force = pressure x area / 150 x 8 = 1200 (N) (2 Marks) b)power = F x v = 1200 x 16 ecf = 19200 (W) (3 Marks) c) (force greater as) air resistance is greater explanation of why: correct quantification, air resistance proportional to v or v2 or in terms of molecules harder collisions or increased rate of collision. power greater as force is greater power greater as velocity is greater (3 Marks) (Marks available: 8) a) F = mg = 6 x 9.81 = 58.86 N (1 Mark) b) Power = Force x distance = Force x speed (1 Mark) time = 58.86 x 3 = 176.58 W (1 Mark) (2 Marks) c) 60 = Power output 100 Power input So power input = Power output x 100/60 (1 Mark) = 176.58 x 100/60 = 294.3 W (1 Mark) (2 Marks) (Marks available: 5)