A lorry moves up a road that is inclined at 9.0° to the horizontal, as shown in Fig. 2.1.

Question 15

(a) (i) Define power. [1]

(ii) Use your definition in (i) to show that power may also be expressed as the product of

force and velocity. [2]

(b) A lorry moves up a road that is inclined at 9.0° to the horizontal, as shown in Fig. 2.1.

Fig. 2.1

The lorry has mass 2500 kg and is travelling at a constant speed of 8.5 m s^-1. The force due to air resistance is negligible.

(i) Calculate the useful power from the engine to move the lorry up the road. [3]

(ii) State two reasons why the rate of change of potential energy of the lorry is equal to the power calculated in (i). [2]

Reference: Past Exam Paper – June 2014 Paper 21 Q2

Solution:

(a) (i) Power is defined as the work (done) per unir time (taken)                             

(ii)

Work done = Force × Displacement (in direction of the force)

So, power = Force × Displacement / time = Force × Velocity

{Displacement / time = velocity}

(b)

(i)

Weight = mg   

{The weight can be resolved into any two components as long as these are perpendicular to each other. For convenience, we will use the components that are easiest to find and more useful in this calculation.

 

The lorry exerts a force along the slope which is equal to the component of the weight along the slope. So, the useful component here is that of the weight along the slope.

It can be deduced that (the basics are done in maths) the angle between the weight and the perpendicular component to the slope is 9°, as indicated in the diagram.

Resolving gives:

Component of weight along slope = mg sin 9°}

Power, P = Fv = (2500 × 9.81 × sin 9°) × 8.5 = 33 (32.6 kW)

(ii)

There is no gain or loss of kinetic energy. {since the speed is constant}      

There is no work (done) against air resistance. {air resistance is negligible}