Physics 9702 Doubts | Help Page 231
Question 1088:
[Work, Energy and Power]
The pump of a water pumping system
uses 2.0 kW of electrical power when raising water. The pumping system lifts 16
kg of water per second through a vertical height of 7.0 m.
What is the efficiency of the
pumping system?
A 1.8% B 5.6% C
22% D 55%
Reference: Past Exam Paper – November 2013 Paper 13
Q17
Solution 1088:
Answer: D.
Input (electrical) power = 2.0 kW
As the water is raised, it gains
gravitational potential energy (= mgh).
Power = Energy / time = mgh / t
The pumping system lifts 16 kg of
water per second. So, (m/t) = 16 kg s-1.
Output Power = (m/t)gh = 16 × 9.81 ×
7.0 = 1098.72 W
Efficiency = Output power / Input
Power = (1098.72 / 2000) × 100% = 55%
Question 1089:
[Dynamics]
(a) Define force.
(b) A resultant force F acts on an object of mass 2.4 kg. The
variation with time t of F is shown in Fig. 2.1.
The object starts from rest.
(i) On Fig. 2.2, show quantitatively
the variation with t of the acceleration a of the object. Include appropriate
values on the y-axis.
(ii) On Fig. 2.3, show
quantitatively the variation with t of the momentum p of the object. Include
appropriate values on the y-axis.
Reference: Past Exam Paper – June 2013 Paper 22 Q2
Solution 1089:
(a) Force is defined as the rate of change of momentum.
(b)
(i)
{Force is constant and
equal to 8.4N from t=0 to t=2. This
corresponds to a constant acceleration of a = F/m = 8.4/2.4 = 3.5ms-2.
Then, at t=2, the resultant force becomes zero and so is the acceleration. For
t=2 to t=4, acceleration remains zero since F is 0.}
Horizontal line on graph from t = 0
to t about 2.0s ± ½ square,
a>0
Horizontal line at a=3.5 on graph
from 0 to 2s
Vertical line at t=2.0s to a=0 or sharp
step without a line
Horizontal line from t=2s to t=4s
with a=0.
(ii)
{Since the object starts
from rest, the initial velocity is 0 at t = 0.
Acceleration, a = 3.5ms-2
for 2s, so [since the object is accelerating from rest for 2s, at time t
= 2s, its velocity is not zero, but is given as follows]
velocity v at t = 2 is v =
u + at = 0 + 3.5(2) = 7ms-1
Momentum, p = mv = 2.4×7 = 16.8Ns. So, a straight line with +ve gradient.
From t=2 onwards, a=0 (but
a = 0 does not mean momentum p is zero since p depends on the velocity, not
acceleration. Acceleration = 0 does not mean that velocity = 0. It only means
that velocity is no changing), so v is constant and so is p. So, straight line (horizontal)
at 16.8Ns.}
Straight line and +ve gradient starting at (0,0) and finishing at (2, 16.8)
Then
a horizontal line from 16.8 for t = 2.0 to 4.0
Question 1090: [Physical Quantities and Units]
The maximum theoretical power P of a wind turbine is given by the
equation
P = kρAvn
where ρ is the density of air, A is the area swept by the turbine
blades, v is the speed of the air and k is a constant with no units.
What is the value of n ?
A 1 B
2 C 3 D 4
Reference: Past Exam Paper – June 2014 Paper 12 Q1
Solution 1090:
Answer: C.
P = kρAvn
vn = P / kρA
Power P = Work done / time = Force ×
distance / time
Consider the units:
[Power P]: [Force F] × [distance s]
/ [time t] = kgms-2 m s-1 = kgm2s-3
k has no units
[kρA]: kgm-3 m2 = kgm-1
[v]: ms-1
So, [vn]: kgm2s-3
/ kgm-1 = m3s-3 = [ms-1]3
Thus, n =3
