# 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 = kρAv

^{n}
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 = kρAv

^{n}
v

^{n}= 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}= kgm^{2}s^{-3}
k has no units

[kρA]: kgm

^{-3}m^{2}= kgm^{-1}
[v]: ms

^{-1}
So, [v

^{n}]: kgm^{2}s^{-3}/ kgm^{-1}= m^{3}s^{-3}= [ms^{-1}]^{3}
Thus, n =3

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