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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