Sunday, November 30, 2014

Physics 9702 Doubts | Help Page 30

  • Physics 9702 Doubts | Help Page 30

Question 183: [Power > Efficiency]
A train on mountain railway is carrying 200 people of average mass 70 kg up slope at angle of 30° to horizontal and at speed of 6.0 m s–1. Train itself has a mass of 80 000 kg. Percentage of power from the engine which is used to raise the passengers and the train is 40 %.
What is power of the engine?
A 1.1 MW                   B 2.8 MW                   C 6.9 MW                   D 14 MW

Reference: Past Exam Paper – June 2013 Paper 12 Q17



Solution 183:
Answer: C.
Total mass in the system = Mass of passengers + Mass of train
Total mass in the system = 200(70) + 80000 = 94000kg

The weight acts vertically downwards. [Weight = mg] Since the slope is at an angle of 30o to the horizontal, the component of the weight acting (down) along the slope is
Component of weight along slope = (94000x9.81) sin30

Power required [output] = Work done / time = Force x velocity

To move at a constant speed of 6.0ms-1, the force provided by the engine should be equal (and opposite) to that component of the weight along the slope, such that the resultant acceleration is zero.
Force required by engine = (94000x9.81) sin30

Power output by engine = Fv = [(94000x9.81) sin30] x 6.0 = 2766420W

Efficiency = Power output / Power input = 40% = 0.4
Power input = 2766420 / 0.40 = 6.9MW









Question 184: [Power]
Diagram shows wheel of circumference 0.30 m. Rope is fastened at one end to force meter. Rope passes over wheel and supports a freely hanging load of 100 N. Wheel is driven by an electric motor at constant rate of 50 revolutions per second.
When wheel is turning at this rate, force meter reads 20 N.

What is output power of the motor?
A 0.3 kW                    B 1.2 kW                     C 1.8 kW                     D 3.8 kW

Reference: Past Exam Paper – June 2013 Paper 11 Q19



Solution 184:
Answer: B.

The force due to the 100N load causes the right end of the rope to accelerate downwards (this simultaneously causes the left end to move upwards). The force due to the 20N at the force meter causes the left end to accelerate downwards (this simultaneously causes the right end to move upwards). Even if both the 100n and the 20N act downwards, they have opposite effects on the rope and the direction of turning of the wheel.
Since the 100N load is greater than the 20N force from the force meter, there is a resultant force acting downwards at the load (that force would cause the wheel to turn in a clockwise direction if not balanced) in the rope.

To keep the load from moving, the wheel should also exert a force equal in magnitude (and in opposite direction) on the rope [from Newton’s 3rd law since the resultant force of the rope is 100 – 20 = 90N downwards].
Force exerted by the wheel on the rope = (100 – 20 =) 80 N


The wheel turns in an anticlockwise direction at a rate of 50 revolutions per second. The distance moved against the force in one second is 50 × 0.30 = 15m.  [Distance moved in 1 second = speed]

Power provided by motor = Fv = 80 N × 15 m s–1 = 1200 W = 1.2kW









Question 185: [Power]
Force of 1000 N is needed to lift hook of a crane at steady velocity. Crane is then used to lift a load of mass 1000 kg at velocity of 0.50 m s–1.
How much of power developed by motor of the crane is used in lifting the hook and the load? Assume that acceleration of free fall g is equal to 10 m s–2.
A 5.0 kW                    B 5.5 kW                     C 20 kW                      D 22 kW

Reference: Past Exam Paper – June 2010 Paper 11 Q15 & Paper 12 Q16 & Paper 13 Q18



Solution 185:
Answer: B.
For steady velocity, the resultant force in the system should be zero (resultant acceleration is zero). So, the force needed to cause the motion of constant velocity should be to equal to the total force in the system.

Total force in the system = force of 1000N needed to lift hook+ weight of the load
Weight of the load = 1000 x 10 = 10000N
Total force needed = 10000 + 1000 = 11000N

Power = Work done / time = Force x distance / time = Force x velocity
Power = (10000 + 1000) x 0.50 = 5.5kW









Question 186: [Matter > Young modulus]
Composite rod is made by attaching glass-reinforced plastic rod and nylon rod end to end, as shown.

Rods have same cross-sectional area and each rod is 1.00 m in length. Young modulus Ep of the plastic is 40 GPa and Young modulus En of the nylon is 2.0 GPa.
Composite rod will break when its total extension reaches 3.0 mm.
What is the greatest tensile stress that can be applied to composite rod before it breaks?
A 7.1 × 10–14Pa
B 7.1 × 10–2Pa
C 5.7 × 106Pa
D 5.7 × 109Pa

Reference: Past Exam Paper – June 2014 Paper 12 Q21



Solution 186:
Answer: C.
The composite rod will break when its total extension reaches 3.0 mm.

For a material,
Young modulus, E = stress, S / strain
Stress = Force / Area
Strain = extension, e / original length, L

E = S / (e/L) = SL / e
Extension, e = SL / E

The same tensile stress (the same force and since the cross-sectional area is the same for both, the same tensile stress is therefore applied) is applied to different materials. The different materials will extend by different amounts when the same stress is applied to each. The total extension is 3.0mm (= 3.0x10-3m).

For the plastic rod, extension ep = S (1) / (40x109) = S / (40x109)
For the nylon rod, extension en = S (1) / (2x109) = S / (2x109)

For the greatest tensile stress that can be applied to composite rod before it breaks,
ep + en = 3.0x10-3  
S/(40x109) + S/(2x109) = 3.0x10-3
Tensile stress, S = (3.0x10-3) / [1/(40x109) + 1/(2x109)] = 5.71x106Pa










Question 187: [Work, Energy, Power]
A ball released from a height h0 above a horizontal surface rebounds to a height h1 after one bounce. The graph that relates h0 to h1 is shown below.

If the ball (of mass m) was dropped from an initial height h and made three bounces, the kinetic energy of the ball immediately after the third impact with the surface was
A (0.8)3 mgh
B (0.8)2 mgh
C 0.8 mg (h/3)
D [1 – (3x0.2)] mgh
E [1 – (0.8)3] mgh

Reference: Past Exam Paper – N81 / II / 2



Solution 187:
Answer: A.
Initially, potential energy of ball = mgh
h0 is the height before release and h1 is the height after one bounce.

Since the graph is a straight line, the ratio of h1 / h0 (gradient) is constant.
Gradient = Δh1 / Δh0 = (80 – 0) / (100 – 0) = 0.8

So, after one bounce, the potential energy becomes 0.8 of the initial potential energy.
The kinetic energy of the ball immediately after an impact with the surface is the same as the potential energy at maximum height [all kinetic energy is converted into potential energy].

After 1st bounce, kinetic energy = 0.8 mgh
After 2nd bounce, kinetic energy = 0.8 [0.8 mgh] = (0.8)2 mgh
After 3rd bounce, kinetic energy = (0.8)3 mgh









Question 188: [Work, Energy, Power]
A mass m moves on a rough plane inclined at an angle θ to the horizontal and, when moving, experiences a constant frictional force F. Mass M is attached to it by means of a light inelastic cord running over a smooth pulley. Mass M is allowed to fall a vertical distance x, causing m to move up the plane as shown in the diagram below.

How much heat is generated by friction in this process?
A Fx
B mgx
C Mgx sinθ
D Mgx sinθ – Fx
E Mgx sin x + Fx

Reference: Past Exam Paper – J88 / I / 6



Solution 188:
Answer: A.
The mass M does not experience any frictional force. Only the mass m, which is on the rough surface, experiences the frictional force F.

When the mass M falls a distance x, mass m also moves a distance x up the plane.  Note that this distance is along the surface, in a direction exactly opposite to the frictional force F vector (the distance moved is not at an angle to the force).

The amount of heat generated by the friction is equal to the work done by the frictional force F. {As the mass m moves, it experiences the frictional force. So, here, the frictional force F acts for a distance x – the distance moved by the mass}

Work done = Force x distance moved in direction of the force
Work done = Fx










Question 189: [Work, Energy, Power]
What is the power required to give a body of mass m a forward acceleration a when it is moving with velocity v up a frictionless track inclined at an angle θ to the horizontal?
A mavg sin θ
B mav sin θ + mgv
C mav + mgv sin θ
D (mav + mgv) sin θ
E (mav + mgv) / sin θ

Reference: Past Exam Paper – N88 / I / 5



Solution 189:
 Answer: C.
Power = Work done / time = Force x velocity

The track is frictionless, so no frictional force acts on the mass.
If the track was horizontal, the power required to give the mass m a forward acceleration a when it is moving with velocity v is given by
Power = Force x velocity = (ma) x v = mav

But in this case, the track is inclined at an angle θ to the horizontal. The component of the weight of the mass (against motion) along the track is mg sin θ. Therefore, a force is required to keep the mass stationary (from moving down along the inclined track). This force should be equal in magnitude to the component of the weight so that the resultant acceleration (when considering only the component of the weight and this force) is zero.
This force would result in a power of mg sin θ x v = mgv sin θ.

Total power required = mav + mgv sin θ










Question 190: [Work, Energy, Power]
The mutual potential energy V of two molecules separated by a distance x is shown in the diagram.

Which of the following correctly describes the force between the molecules?

Reference: Past Exam Paper – J89 / I / 6



Solution 190:
Answer: E.
There are 2 forces between the two molecules. One force is attractive and predominates when the molecules are far apart and the other force is repulsive and predominates when the molecules are close to each other.
So, there is a particular separation of the molecules when the attractive and repulsive forces balance. That is, there is a separation when the resultant force between the molecules is zero.

The force, F is related to the potential energy, V by
F = - dV / dr

Thus, when the force is zero, dV / dr is zero. That is, the gradient of a potential energy, V against separation r graph is zero. This occurs where the graph is either a maximum or a minimum. This point is indicated by r2 in the question.

So, for separations less than r2, the force between the molecules is repulsive and for separations greater than r2, the force is attractive.









Question 191: [Work, Energy, Power]
An object of mass m passes a point X with a velocity v and slides up a frictionless incline to stop at point Y which is at a height h above X.

A second object of mass ½ m passes X with a velocity of ½ v. To what height will it rise?
A ¼ h              B ½ h              C (1/√2) h                    D h                  E h√2

Reference: Past Exam Paper – J87 / I / 5



Solution 191:
Answer: A.
For the object of mass m and velocity v,
At point X, Kinetic energy = ½ mv2

From conservation of energy,
Potential energy at point Y = Kinetic energy at point X
mgh = ½ mv2  
Height, h = v2 / 2g


For the object of mass m / 2 and velocity v /2,
At point X, Kinetic energy = ½ (m/2) (v/2)2 = mv2 / 16

Potential energy at point Y = Kinetic energy at point X
Let the height at point Y = x
(m/2)gx = mv2 / 16
Height, x = v2 / 8g = (v2 / 2g) / 4 = h / 4










Question 192: [Work, Energy, Power]
The diagram shows two bodies X and Y connected by a light cord passing over alight, free-running pulley. X starts from rest and moves on a smooth plane inclined at 30o to the horizontal.

What will be the total kinetic energy of the system when X has travelled 2.0m along the plane? (g = 9.8ms-2)
A 20J               B 59J               C 64J               D 132J             E 137J

Reference: Past Exam Paper – J90 / I / 7



Solution 192:
Answer: B.
There is an acceleration (due to gravity) of 9.8ms-2 downwards on Y since it is hanging freely.
Weight of Y = mg = 5.0 x 9.8 = 49N

However, this does not only cause Y to move. X, which is attached to Y will also move. So, the acceleration in the whole system is therefore NOT 9.8ms-2. A resultant acceleration (which acts on the total mass in the system) should be calculated.

It should also be noted that the component of the weight of X will also act on the system. This component acts (downward) along the plane and opposes the direction of motion of X (X should move 2.0 [up] along the slope).
Component of weight of X along plane = mg sin(30) = 4.0 x 9.8 x sin(30) = 19.6N

{The weight of Y contributes a force (up) along the slope while the component of weight of X contributes a force (down) along the slope}
Resultant force in system = ma = 49 – 19.6
(4.0 + 5.0) a = 29.4N
Acceleration, a = 29.4 / 9.0 = 3.27ms-2
This is the resultant acceleration on each X and Y.


Initially both X and Y are at rest.
Consider the equation for uniformly accelerated motion: v2 = u2 + 2as
Initial speed, u = 0
Final speed = v
Distance travelled, s = 2.0m
Acceleration, a = 3.27ms-2

v2 = 02 + 2(3.27)(2.0)
Final speed, v = √[2(3.27)(2.0)] = 3.62ms-1

Kinetic energy of X = ½ (4.0) (3.62)2
Kinetic energy of Y = ½ (5.0) (3.62)2
Total kinetic energy of system = ½ (4.0) (3.62)2 + ½ (5.0) (3.62)2 = 58.96 = 59N    





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