Exam-style Questions: Power and Internal Energy

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 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)