# Exam-style Questions: Power and Internal Energy

1. 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)

2. 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:

1. 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)

2. 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)