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}
