Tuesday, December 9, 2014

Physics 9702 Doubts | Help Page 33

  • Physics 9702 Doubts | Help Page 33

Question 204: [Work, Energy, Power]
(a) Starting with the definition of work, deduce the change in the gravitational potential energy of a mass m, when moved a distance h upwards against a gravitational field of field strength g.

(b) By using the equations of motion, show that the kinetic energy EK of an object m travelling with speed v is given by EK = ½ mv2.

(c) A cyclist, together with his bicycle, has a total mass of 90kg and is travelling with a constant speed of 15ms-1 on a flat road at A, as illustrated in Fig. He then goes down a small slope to B so descending 4.0m.

Calculate
(i) the kinetic energy at A
(ii) the loss of potential energy between A and B
(iii) the speed at B, assuming that all the lost potential energy is transformed into kinetic energy of the cyclist and bicycle.

(d)
(i) A cyclist travelling at a constant speed of 15ms-1 on a level road provides a power of 240W. Calculate the total resistive force.
(ii) The cyclist now travels at a higher constant speed. Explain why the cyclist needs to provide a greater power.

(e) It is often stated that many forms of transport transform chemical energy into kinetic energy. Explain why a cyclist travelling at constant speed is not making this transformation. Explain what transformations of energy are taking place.

Reference: Past Exam Paper – N99 / III / 1



Solution 204:
(a)
Work done = Force x distance moved in direction of force
(Gravitational) Force acting on mass m = Weight = mg

The upward force required to move the mass m a distance h against the gravitational field is equal (in magnitude) to mg (it is opposite in direction since the weight acts downwards while this force acts upwards).

From the law of conservation of energy, the gain in the gravitational potential energy is equal to the work done.
Change in gravitational potential energy = mgh

(b)
Work done = Force x distance moved in direction of force
Force = mass x acceleration = ma
Distance moved in direction of force = s
Work done = ma s

EITHER
Distance travelled = average speed x time = [(v + u) / 2] t
where v is the final speed and u is the initial speed

Acceleration, a = (v – u) / t

Work done = m {(v – u) / t} {[(v + u) / 2] t} = ½ m (v2 – u2) = ½mv2 – ½mu2

Let the initial speed, u = 0
Kinetic energy = work done = ½ mv2


OR
Consider the equation of uniformly accelerated motion: v2 = u2 + 2as
as = (v2 – u2) / 2

Work done = m {(v2 – u2) / 2} = ½ mv2 – ½ mu2
Each of the 2 terms, ½ mv2 and ½ mu2, represent the final and initial kinetic energy respectively.

So, kinetic energy at speed v = ½ mv2

(c)
(i)
Kinetic energy at A = ½ mv2 = 0.5 (90) (15)2 = 10125kJ = 10.1kJ

(ii)
Loss of potential energy between A and B = mgh = 90 (9.81) (4.0) = 3.53kJ  

(iii)
Total Kinetic energy at B = kinetic energy at A + loss in potential energy between A and B
Total Kinetic energy at B = 10125 + 3531.6 = 13656.6J
½ mv2 = 13656.6
Speed at B, v = √[(2 x 13656.6) / 90] = 17.4ms-1

(d)
(i)
For constant speed, resultant acceleration and hence resultant force must be force. So, the force provided should be equal to the total resistive force.
Power P = Fv
Resistive force F = P / v = 240 / 15 = 15N

(ii)
Power P = Fv. For a higher constant speed (v is greater), the power also needs to be higher. Additionally, at a higher speed, the air resistance increases {Air resistance: F = kv2} So, F is greater, so power P should also be greater.

(e)
For many forms of transport, the chemical energy is from the fuel being burnt. A bicycle does not use any fuel, so the transformation is not from chemical energy to kinetic energy.
For a cyclist, the energy comes from the cyclist himself. From the breakdown of food in his body, the chemical energy available is transformed into mechanical energy as the cyclist is pedaling. Some of the energy is turned into thermal energy in the muscles, the tires and the bearings. The mechanical energy is then transformed into (rotational) kinetic energy as the tires move. Some energy turns into frictional loss.










Question 205: [Matter > Elastic and Plastic behavior]
Graph shows behaviour of a sample of a metal when it is stretched until it starts to undergo plastic deformation.

What is the total work done in stretching sample from zero extension to 12.0 mm?
Simplify calculation by treating the region XY as a straight line.
A 3.30 J           B 3.55 J           C 3.60 J           D 6.60 J

Reference: Past Exam Paper – June 2006 Paper 1 Q22 & November 2014 Paper 13 Q24



Solution 205:
Answer: B.
Work done = force x distance = area under graph
The area under graph can be considered by first dividing the area into a triangle and a trapezium.

Area of triangle = 0.5 (500) (0.010) = 2.50J
Area of trapezium = 0.5 (500 + 550) (0.002) = 1.05J
Total area = 2.50 + 1.05 = 3.55J

3 significant figures are possible here.









Question 206: [Forces > Equilibrium]
A heavy uniform beam of length l is supported by two vertical cords as shown.

What is the ration of T1 / T2 of the tensions in these cords?
A 1 / 3             B 1 / 2             C 2 / 1             D 3 / 1

Reference: Past Exam Paper – N92 / I / 5 & J95 / I / 5



Solution 206:
Answer: A.
The beam is uniform, so its weight acts downwards at the middle.
Let the weight of the beam be W.

Let the left end of the beam {where the tension is T1 in the cord} act as the pivot. {Note that the point where the other point is attached could have alternatively been taken as the pivot and T1 is found first. The calculations would be different but the final ratio would be the same}

The weight provide a clockwise moment of (0.5L) W = 0.5LW
The tension T2 in the cord provides an anticlockwise moment on the beam.

For equilibrium, the clockwise moment on the beam should be equal to the anticlockwise moment.
(2L / 3) T2 = 0.5LW
Tension T2 = 3W / 4

For equilibrium, the resultant force on the beam should also be zero.
Upward force = Downward force
T1 + T2 = W
T1 + 3W / 4 = W
Tension T1 = W / 4

Ratio = T1 / T2 = (W/4) / (3W/4) = 1 / 3









Question 207: [Forces > Equilibrium]
A rod of length 1 metre has non-uniform has non-uniform composition, so that the centre of gravity is not at its geometrical centre.
The rod is laid on supports across two top-pan balances as shown in the diagram. The balances (previously set at zero) give readings of 360g and 240g.

Where is the centre of gravity of the rod relative to its geometrical centre?
A 1 / 10 metre to the left
B 1 / 10 metre to the right
C 1 / 6 metre to the elft
D 1 / 5 metre to the right
E 1 / 5 metre to the left

Reference: Past Exam Paper – J93 / I / 5



Solution 207:
Answer: A.
For the rod to be horizontal (in equilibrium), the net moment on it should be zero.

Consider the moments about the right support.
Let the centre of gravity be at a distance x from the right support and let the acceleration of free fall be a {so as not to confuse with g in gram}.
Total mass of the rod = 360 + 240 = 600g

The weight of the rod (due to its centre of gravity) provides an anticlockwise moment of 600gx Nm.

From Newton’s third law, the support on the left (which is at a distance of 1m from the right support) will provide an upward force equal to the force acting downwards on it. So, the left support provides a clockwise moment of 360g(1) = 360g Nm.

For equilibrium,
600gx = 360g
Distance x from right support = 360 / 600 = 0.6m

The centre of gravity is at a distance of 0.6m from the right end. Therefore, it is (0.6 – 0.5 =) to the left of the left of the geometrical centre (which is at 0.5m from each end of the rod).









Question 208: [Forces > Moment]
Uniform metre rule of weight 2.0N is pivoted at 60cm mark. A 4.0N weight is suspended from one end, causing the rule to rotate about the pivot.

At the instant when rule is horizontal, what is the resultant turning moment about the pivot?
A zero                         B 1.4 N m                   C 1.6 N m                   D 1.8 N m

Reference: Past Exam Paper – N94 / I / 5 & November 2013 Paper 11 & 12 Q15



Solution 208:
Answer: B.
Since the metre rule is uniform, its weight act at the 50cm mark (that is, at a distance of 10cm from the pivot). This produces an anticlockwise moment of (2.0 x [0.6 – 0.5]).

The weight of 4.0N acts downwards at one end, it produces a clockwise moment of (4.0 x (100-60) x 10-2).

At equilibrium, the rule will always remain horizontal but this is not the case here. We are talking about an INSTANT in this question. So, resultant turning moment cannot be zero.

Resultant turning moment = (4.0 x (100-60) x 10-2) – (2.0 x 0.1) = 1.4Nm










Question 209: [Forces > Equilibrium]
The force diagrams show all the forces acting on a beam of length 3x.
Which force system causes only rotational motion of the beam without any linear movement?


Reference: Past Exam Paper – N98 / I / 5



Solution 209:
Answer: D.
Linear movement is caused when the resultant force in the system is not zero. Rotational motion results whenever the resultant moment ion the system is not zero.

Therefore, in this question, the resultant force should be zero while the resultant moment (torque) is non-zero.
For the resultant force to be zero, the sum of upward forces (FUP) should be equal to the sum of downward forces (FDOWN). For analysis of the moments of the forces, the moments of the forces will be taken about the left end of the beam. The downward forces would then cause a clockwise moment while the upward forces would cause an anticlockwise moment. Additionally, any force acting at the left end of the beam would not produce any moment since its ‘distance from the pivot would be zero’.

Consider A:
Resultant force = FUP – FDOWN = (2+5) – (3+4) = 0

Clockwise moment = 4(3x) = 12x Nm
Anticlockwise moment = 2(x) + 5(2x) = 12x Nm
The resultant moment is zero, so no rotational motion of the beam occurs.

Consider B:
Resultant force = FUP – FDOWN = (5+3) – (4+2) = 2N
The resultant force is not zero, so there is linear movement of the beam.

Consider C:
Resultant force = FUP – FDOWN = 5 – (2+3+4) = 4N
The resultant force is not zero, so there is linear movement of the beam.

Consider D:
Resultant force = FUP – FDOWN = (4+3) – (2+5) = 0

Clockwise moment = 5(2x) = 10x Nm
Anticlockwise moment = 4(x) + 3(3x) = 13x Nm
 The resultant moment is not zero, so there is rotational motion of the beam.





10 comments:

  1. in q208, why is the answer not A(i.e. zero)?
    It should be A as the question reads that what is the moment about the pivot when it is "horizontal", and the rule can only be horizontal when the net moment is zero.

    ReplyDelete
    Replies
    1. If the resultant turning moment is zero, then the clockwise moment should be equal to the anticlockwise moment. This would imply equilibrium BUT this is obviously not the case (calculate each moment and you'll find out).

      The rule is horizontal only for an INSTANT. Afterwards, it's gonna rotate. We are calculating the resultant turning moment at that particular moment.

      Delete
  2. In question 205 the answer we get is 3.55 j but in reality the area is a little more than that isn't it ? since we approximated XY to a straight line so shouldn't the answer be 3.60 j

    ReplyDelete
    Replies
    1. The question itself said to approximate the region XY as a straight line, so there's no need to find out the actual value if it was a curve.

      + we cannot be sure whether that small account that has not been accounted for would correspond to 0.05J.

      Delete
  3. Hello, for question 204 c part 3 why is it that " Total Kinetic energy at B = kinetic energy at A + loss in potential energy between A and B" ? Cant i just use loss in PE between A and B=KE at B?

    ReplyDelete
    Replies
    1. No, because the KE at A is not zero since the cyclist has some speed. Energy cannot be destroyed and energy is always conserved - so we need to account for it too.

      Delete
    2. I see, thank you very much!!

      Delete
  4. Can you explain question 206 again ?? i didnt get that

    ReplyDelete
    Replies
    1. tell me exactly what you did not understand

      Delete

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