• Physics 9702 Doubts | Help Page 16

Question 85: [Forces > Resultant force]
Car of mass m travels at constant speed up slope at an angle θ to horizontal, as shown in diagram. Air resistance and friction provide resistive force F.

What force is needed to propel car at this constant speed?
A mg cos θ                  B mg sin θ                   C mg cos θ + F            D mg sin θ + F

Reference: Past Exam Paper – June 2012 Paper 13 Q15



Solution 85:
Answer: D.
For constant speed, the forward force should be equal to the total backward force so that the resultant acceleration is zero.
Component of weight along slope = mg sin θ
Total backward force = mg sin θ + F

Forward force needed = Total backward force = mg sin θ + F








Question 86: [Waves > Intensity]
Sound wave consists of series of moving pressure variations from normal, constant air pressure.
Graph shows these pressure variations for two waves at one instant in time.

Wave 1 has intensity of 1.6 × 10^–6W m^–2.
What is intensity of wave 2?
A 2.4 × 10^–6W m^–2      B 3.0 × 10^–6W m^–2      C 3.6 × 10^–6W m^–2      D 4.5 × 10^–6W m^–2

Reference: Past Exam Paper – June 2014 Paper 12 Q23



Solution 86:
Answer: C.
Intensity is proportional to amplitude square, A².

For wave 1, amplitude A = 2 (from graph).
So, 2² = 4[units] corresponds to the intensity of 1.6x10^-6Wm^-2.

Wave 2, which has amplitude A = 3 (from graph), will have intensity corresponding to 3² = 9[units].
This (9 units) is equal to an intensity of (9/4) x 1.6x10^-6 = 3.6x10^-6Wm^-2








Question 87: [Resistance > Power]
Diagram shows low-voltage circuit for heating water in a fish tank.

Heater has resistance of 3.0 Ω. Power supply has e.m.f. of 12 V and internal resistance of 1.0 Ω.
At which rate is energy supplied to heater?
A 27 W                       B 36 W                        C 48 W                        D 64 W

Reference: Past Exam Paper – June 2014 Paper 13 Q35



Solution 87:
Answer: A.
Rate of energy supplied to heater = power in heater
Power in heater = I²R                       where R is the resistance of the heater.

Total resistance = 3.0 + 1.0 = 4.0Ω
Current = emf / total resistance = 12 / 4 = 3.0A

Power in heater = (3²) (3) = 27W








Question 88: [Forces > Torque]
Four beams of the same length each have three forces acting on them.
Which beam has both zero resultant force and zero resultant torque acting?

Reference: Past Exam Paper – November 2012 Paper 13 Q16 & November 2016 Paper 11 & 13 Q15



Solution 88:
Answer: D.
Torque is the moment of a torque. It causes a turning effect.

Consider A:
Upward force = 30 + 50 = 80N
Downward force = 90N
Resultant force = 90 – 80 = 10N downward
Resultant force is not zero. So, A is incorrect

Consider B:
Upward force = 36 + 70 = 106N
Downward force = 106N
Resultant force = 106 – 106 = 0

Consider forces acting about the centre of mass (where the 106N acts downwards),
Clockwise moment = 36 (50) = 1800Ncm
Anticlockwise moment = 70 (30) = 2100Ncm
Resultant moment = 2100 – 1800 = 300Ncm
Resultant moment is not zero. So, B is incorrect

Consider C:
Upward force = 28 + 35 = 63N
Downward force = 63N
Resultant force = 63 – 63 = 0

Consider forces acting about the centre of mass (where the 63N acts downwards),
Clockwise moment = 28 (50) = 1400Ncm
Anticlockwise moment = 35 (30) = 1050Ncm
Resultant moment = 1400 – 1050 = 350Ncm
Resultant moment is not zero. So, C is incorrect

Consider D:
Upward force = 42 + 70 = 112N
Downward force = 112N
Resultant force = 112 – 112 = 0

Consider forces acting about the centre of mass (where the 112N acts downwards),
Clockwise moment = 42 (50) = 2100Ncm
Anticlockwise moment = 70 (30) = 2100Ncm
Resultant moment = 2100 – 2100 = 0
Both the resultant force and the resultant moment are zero. So, D is correct









Question 89: [Forces > Torque]
Two 8.0N forces act at each end of beam of length 0.60m. Forces are parallel and act in opposite directions. Angle between forces and beam is 60 °.

What is the torque of the couple exerted on beam?
A 2.4 N m                   B 4.2 N m                   C 4.8 N m                   D 9.6 N m

Reference: Past Exam Paper – June 2007 Paper 1 Q13



Solution 89:
Answer: B.
A couple consists of 2 forces, equal in magnitude and opposite in direction, acting on different points such that they produce a turning effect.

Torque of the couple, τ = F x d
where F is the magnitude of one of the forces and d is the perpendicular distance between the 2 forces

 {We cannot use the distance as 0.60m since this is NOT the perpendicular distance between the forces. + There is no need to identify any pivot since perpendicular distance between the 2 forces is required.}

{Now, here is a working that was suggested (it is not a correct method to solve the question).
Consider the centre of the beam and try to calculate the corresponding length that would be perpendicular to the forces and pass through the origin. Let the perpendicular distance from the centre to 1 of the forces = x

x = 0.30 sin60
So, the total ‘resolved’ perpendicular distance between the 2 forces = 2 (0.30 sin60)
Torque, τ = F x d = 8.0 x 2 (0.30 sin60) = 4.156 = 4.2Nm

Even if the correct answer is obtained, this method is ‘physically’ incorrect because the beam is a ‘physical’ material with a ‘physical’ length, so we cannot resolve for another length that does not actually exist.

The correct what would be to resolve for the perpendicular (vertical) component of the forces. This is accepted since force is a concept (and this method is much easier and quicker). This is what is described next.}

Component of one force perpendicular to beam = 8sin60
Torque = 0.6 x 8sin60 = 4.157 = 4.2Nm