Question 27
(a)
(i) Define power. [1]
(ii)
State what is meant by gravitational potential energy.
[1]
(b)
An aircraft of mass 1200 kg climbs upwards with a constant velocity
of 45 m s-1, as shown in Fig. 3.1.
Fig. 3.1 (not to scale)
The aircraft’s engine
produces a thrust force of 2.0 × 103 N
to move the aircraft through the air. The rate of increase in height of the
aircraft is 3.3 m s-1.
(i)
Calculate the power produced by the thrust force. [2]
(ii)
Determine, for a time interval of 3.0 minutes,
1.
the work done by the thrust force to move the aircraft, [2]
2.
the increase in gravitational potential energy of the aircraft, [2]
3.
the work done against air resistance. [1]
(iii)
Use your answer in (b)(ii) part 3 to
calculate the force due to air resistance acting on the aircraft. [1]
(iv)
With reference to the motion of the aircraft, state and explain whether
the aircraft is in equilibrium. [2]
[Total: 12]
Reference: Past Exam Paper – November 2018 Paper 21 Q3
Solution:
(a)
(i) Power is defined as the rate of doing work.
(ii)
Gravitational potential energy is the energy possessed
by a mass due to its position in a gravitational field.
(b)
(i)
P = Fv
{We need to use velocity of the aircraft, which
is in the direction of the thrust (i.e. v = 45 m s-1), not the rate
of increase in height (i.e. not 3.3 m s-1).}
P = 2.0 × 103 × 45
P = 9.0 × 104 W
(ii)
1.
{Power
= work done / time
Work
done = F × d or Work done = Power
× time
Distance d = Speed × time = (45 × 3.0 × 60)}
W = (2.0 × 103)
× (45 × 3.0 × 60)
or W = 9.0×104 × 3.0 × 60
W = 1.6 × 107 J
2.
(Δ)EP = mg(Δ)h
{(Δ)h is the increase in height.
Rate of increase in height (= speed) = 3.3 m s-1
Speed = distance / time So, Distance = Speed × time
Distance (increase in height) = 3.3 × 3.0×60
m}
(Δ)EP =
1200 × 9.81 × 3.3 × 3.0
× 60
(Δ)EP =
7.0 × 106 J
3.
{Work
done by thrust force = Gain in GPE + Work done against air resistance
Work
done against air resistance = Work done by thrust – Gain in GPE}
W = 1.6×107 – 7.0×106
W = 9.0 × 106 J
(iii)
{Against air resistance,
Work done = Force × distance moved in direction
of force
Distance moved in direction of force = Speed × time = (45 × 3.0 × 60)
m}
Force = (9.0 × 106) / (45 × 3.0
× 60)
F = 1.1 × 103 N
No comments:
Post a Comment
If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation