• 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

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• Circular motion in a vertical plane

Consider a pail of water whirling at the end of a rope in a vertical circle. The tension T in the rope varies with the position of the pail.

The figure below shows the forces acting on the pail when the rope is at an angle of θ to the vertical.

From Newton’s 2^nd law: Resultant force F = ma

Net force towards the centre of the circle O is

T – mg cosθ

The net force provides the centripetal acceleration.

Therefore, T – mg cosθ = mv² / r

where v = speed of pail; r = radius of circle

Tension T = mv²/r + mg cosθ

Now, let’s consider how the tension changes as the position of the pail changes.

·         When the pail is at the lowest point (θ = 0),

Tension T = mv²/r + mg        (which is the maximum tension)

·         When the rope is horizontal (θ = π/2 (=90°) or θ = 3π/2 (=270°)), cosθ = 0 and thus

Tension T = mv²/r

·         When the pail is at the highest point (θ = π = 180°), cosθ = –1 and

Tension T = mv²/r – mg        (which is the minimum tension)

Now if the pail reaches the highest point (at a distance equal to the radius r form centre O of the circle), the rope must be taut (there should be a tension in the rope, which is not zero).

( T = mv²/r – mg )      > 0

mv²/r > mg

v > √(gr)

Thus, they velocity of the pail at the highest point must be greater than √(gr).

Now consider the forces acting on the water in the pail when it is at the highest point.

For the water not to fall out (that is, the water is in contact with the pail – there is a contact force on the pail), there should be a reaction R (from Newton’s 3^rd law) on the water by the base of the pail. The resultant force towards the centre of the circle is

F = R + mg = mv² / r

The centripetal force is provided by the reaction R and the weight mg.

R = mv²/r – mg

For the water not to fall, R > 0 (that is, there should be a reaction R on the water).

( R = mv²/r – mg )      > 0

mv² / r > mg

Therefore, water in the pail does not fall out because the centripetal force is greater than the weight of the water. (The weight mg of the water is less than the required force mv²/r towards the centre and so the water stays in. The reaction R of the bucket base on the water provides the rest of the force mv²/r.)

Remember that the weight acts vertically downwards (towards the ground). The centripetal force mv²/r acts towards the centre, causing the bucket to move in a circle.

On the other hand, if the pail is whirled slowly, then mg > mv²/r. The weight is greater than the centripetal force and the water falls out of the bucket. (Here, part of the weight provides the force mv²/r. The rest of the weight causes the water to accelerate downward and hence to leave the bucket.)

References:

1. “Pacific Physics A Level,” by POH LIONG YONG, pg 154 – 155 ‘Motion in a vertical circle’

2. “Advanced Level Physics,” by Nelkon & Parker, 7^th edition, pg 63 – 64