Examstyle Questions: Power and Internal Energy

The front of a lorry can be considered a flat vertical surface of area 8.0m^{2}. 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 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 v^{2} 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)