# Exam-style Questions: Work, Energy and Efficiency

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. Fig. 4.1 shows part of a fairground ride with a carriage on rails. The carriage of mass 500 kg is travelling towards a slope inclined at 30° to the horizontal. The carriage has a kinetic energy of 25 kJ at the bottom of the slope. The carriage comes to rest after travelling up the slope to a vertical height of 3.9 m.

a) Show that the potential energy gained by the carriage is 19 kJ.

(2 Marks)

b) Calculate the work done against the resistive forces as the carriage moves up the slope.

work done = ................. kJ

(1 Mark)

c) Calculate the resistive force acting against the carriage as it moves up the slope.

resistive force = ............................. N

(3 Marks)

(Marks available: 6)

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) potential energy = mgh / weight x height

= 500 x 9.81 x 3.9

= 19. (130) (kJ).

(2 Marks)

b) work done = 25 - 19

= 6 (5.870) (kJ)

(1 Mark)

c) distance up the slope = 7.8m

work done = force x distance

force = 5870 / 7.8

= 753 (N) (769 N if 6000 J is used)

(3 Marks)

(Marks available: 6) 