y12 Sankey Diagrams

Constructing Sankey Diagrams








1. An energy efficient light bulb is rated at 20W. It produces 5W of light. Calculate its efficiency and draw a Sankey Diagram to scale.



2. Paula transfers 40 000J of chemical energy during a race. She transfers 32 000J of heat energy to the surroundings during the race. Calculate her efficiency and draw a Sankey diagram to scale.



3. Bradley does 1600J of work turning the pedals on his bike. 1577.6J is transferred to the rear sprockets. How much heat is lost and what is the efficiency of Bradley’s bike. Draw a Sankey diagram of Bradley’s chain



4. The coal in Thomas’ boiler contains 40 kJ of energy. He loses 25.76J as puffs along a branch line. How efficient is Thomas and draw a Sankey diagram.



5. The fuel in Diesel’s tank contains 60kJ of chemical energy. He does 20.1kJ of work on the mainline. Calculate Diesel’s efficiency and draw a Sankey diagram.



6. Jeremy is the proud owner of an electric car. The electric battery in Jeremy’s car can develop 3kW. If the engine develops 2.658kW what is its efficiency.



7. Jeremy dreams of a Ferrari Enzo which can develop a maximum of 700 bhp. (1 bhp = 750 Watts). Sadly for Jeremy petrol cars are not very efficient. Typically, only about 30% of the energy that is available from the combustion of the petrol actually ends up overcoming friction to move the car forwards. Of the 70% of energy is that is not usefully converted, 55% may heat the cooling water that surrounds the engine block whilst 15% may be in the hot exhaust gases. To make car engines more efficient the fuel has to burn at a higher temperature and the exhaust must be kept cooler. Draw a Sankey diagram of Jeremy’s dinosaur.

Y12 Newton's Second Law

Newton’s Second Law of Motion
(Use g = 9.8 N kg^-1)

Worked example

Find the acceleration of a body of mass 10kg when it is subjected to a horizontal force of 100 N if it (a) can move along a smooth horizontal surface, (b) can move along a horizontal surface which produces a frictional force of 80N.
(a) F=ma      F force in N, mass in kg, a acceleration in ms^-2
Rearranging  a = F/m        a = 100 / 10  = 10 ms^-2

(b) The resultant force = 100 N – 80 N = 20 N

F=ma rearranging a = F/m
 a = 20 /10   a = = 2ms^-2  

1. A force of 100 N acts on a mass of 1 kg. What is the acceleration?
2. A mass of 10kg acquires a velocity of 20 m s from rest in 4 s. Calculate the force is required.
3. A rocket of mass 800 000 kg has motors giving a thrust of 9 800 000 N. Calculate the acceleration at lift off.
4. A force of 5 N acts on a stationary mass of 2kg which can move along a smooth horizontal surface. Calculate its velocity after 5 s?
5. A car of mass 600 kg travelling at 72km h^-1 is brought to rest in 54 m after the driver sees an obstruction ahead. If the distance travelled after the driver applies the brakes is 40m calculate the driver’s reaction time and the braking force.
6. A mass of 2kg projected along a flat surface with a velocity of 15ms^-1 comes to rest after travelling 30 m. Calculate the frictional force.
7. A Mini of mass 576 kg can accelerate from rest to 72 km h^-1 in 20 s. If the acceleration is assumed uniform calculate this acceleration and the tractive force in Newtons needed to produce it.
8. A Mini of mass 576 kg can be stopped (in neutral) in 72 m from 108 kmh^-1. Calculate (a) the deceleration, (b) the frictional force between the tyres and the road in Newtons.
9. The first-stage rocket motors of the Apollo spacecraft produce a thrust of 3.3 x 10⁷ N and the complete spacecraft has a mass of 2.7 x 10⁶ kg. Calculate (a) the resultant force accelerating the spacecraft, (b) the initial acceleration, (c) the time for it to rise through a distance equal to its own height as it ‘lifts off’ if its height is 111 m and the average acceleration during this time is 2.5 ms^-2
10. A boy of mass 50kg stands in a lift. What will he ‘weigh’ in Newtons if the lift accelerates at 0.50 ms^-2 (a) upwards, (b) downwards?
    
    

 

y12 Force

Force

F = force (Newtons)       m = mass (kg)               a= acceleration (ms-2)

 

 

 

1. A car hits a wall and decelerates at 40 ms^-2.  It exerts a force of 20 000 N on the wall.  Calculate the mass of the car?

 

2. A stone of mass 0.5 kg falls into a sandpit exerting a force of 5N.  Calculate its acceleration?

 

3. A car of mass 1000 kg travelling at 10 ms^-1 stops in 4 s. Calculate a) the average deceleration. b) the braking force.

 

4. An aeroplane accelerates uniformly from rest (0) to 200 ms^-1 in 40 s. Calculate the force it exerted on a pilot whose mass is 90 kg?

 

5. A bullet of mass 20 g travelling with a velocity of 300 ms^-1 hits an earth bank and is brought to rest in 0.2 s Find      a) the deceleration of the bullet. b) the force it exerts on the earth bank.

 

6. A car of mass 750 kg stops uniformly in 5s.  If its initial velocity was 15 ms^-1. Calculate the braking force.

 

7. The valve of a cylinder on an engine containing 12 kg of compressed gas is opened and the cylinder empties in 1min 40 sec. If the gas leaves its chamber at 25 ms^-1.Calculate (a) the acceleration of the gas. (b) the force exerted on the chamber.

y13 Circular Motion answers

I have left the answers for the question on Magnetic fileds as it will be relevant when you have studied unit 5

1.       (i)      (v = 2πr/t) t = 2π60/0.26 = 1450 s

Correct answer is 1449.96 hence allow 1.4 × 103 
Do not allow a bare 1.5 × 103

B1

  (ii)   correct substitution into F = mv2/r: eg F = (9.7 × 103 × 0.262)/60

C1

          F = 10.9 N

Allow 11 N

A1

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  2.     (i)      THREE correct arrows at A, B and C all pointing towards
the centre (judged by eye)

Ignore starting point of arrow

B1

  (ii)   1.      Greatest reaction force is at C

This is a mandatory M mark. The second mark cannot be gained unless this is scored.

M1

         because it supports weight of sock AND provides the required
upward resultant (centripetal) force (WTTE)

Any indication that candidates think that the centripetal force is a third force loses this second and possibly the next mark. They must make correct reference to the resultant force that provides the required centripetal force/acceleration.

A1

2.      Least at A because sock’s weight provides part of the required
downward resultant (centripetal) force (WTTE)

Allow answers using the equation F = mv2/r
such as Nc – mg (at C) = centripetal force OR mv2/r
OR mg +NA (at A) = centripetal force OR mv2/r

B1

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  3.     (i)      At top of loop, the centripetal force = mv2 / r (1)
= mg (1)
Thus speed at top v = √gr
= √9.81 × 9.17 / 2 (1)
= 6.7 m s–1                                                                                                       3

          or use an energy argument KE on entry = PE gained + KE at top

          (1 mark for idea, 1 mark for correct substitution in appropriate formula
and 1 mark for correct calculation of 6.7 m s–1)

          (NOTE The unexplained use of v2 = u2 + 2gs can only score a
maximum of 1 mark)

(ii)     Kinetic Energy at top = ½ × 86 × 6.72 (1)
= 1935 J (1)
Potential Energy at top = 86 × 9.81 × 9.17 (1)
= 7740 J (1)                                                                                                      4

  (iii)  Kinetic Energy on entry = ½ × 86 × 152
= 9675 J (1)
Sum of energies at top = 1935 + 7740
= 9675 J .......... QED (1)                                                                                  2

  (iv)  Any reference to loss of contact / centripetal force or wtte (1)
Comment on the consequences of taking off vertically or wtte (1)                  2

          Enacting the suggestion could result in disaster
At the point A in the loop, the velocity vector is purely vertical
Therefore there is no horizontal component of velocity
So no matter how fast the cyclist is travelling he will only be
projected vertically
And come (crashing?) down on the same point where he left off
The best that could happen ( with some skill ) is to return back along same path

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  4.     (a)     B = F/Il with symbols explained or appropriate statement in words; (1)
explicit reference to I and B at right angles/define from F = BQv etc (1)          2

  (b)   (i)      arrow towards centre of circle                                                                 1

(ii)     field out of paper; Fleming’s L.H. rule/moving protons act as
conventional current                                                                               2

  (iii)  F = Bev allow BQv                                                                                  1

  (iv)  F = mv2/r; Bev = mv2/r; (2)
B = mv/er = 1.67 × 10–27 × 1.5 × 107/(1.6 × 10–19 × 60); = 0.0026; T (3) 5

allow Wb m–2

  (v)   the field must be doubled; (1)
B ∞ v (as m, e and r are fixed)/an increased force is required
to maintain the same radius (1)                                                               2

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  5.     (a)     (i)      speed v = 2π r / t
v = 2 × π × 122/2 /(30 × 60) (1)
v = 0.21 m s–1 (1) allow 0.2 m s–1                                                           2

  (ii)   F = 12.5 kN × 16 = 200 kN (1)                                                                1

(iii)    W = F × s or
= 200 k × 2 × π × 122 / 2 (1) ecf (ii) allow ecf for distance from (i)
= 7.7 × 107 J (1) allow 8 × 107                                                                2

  (iv)  P = W / t, energy / time or F × v or
= 7.67 × 107 / (30 × 60) (1) or ecf (iii) / (30 × 60)
= 42.6 kW (1) allow 43 kW, only allow 40 kW if working shown             2

(v)     •        Friction force at bearing opposes motion so not useful (1)
•        Friction force of tyres on rim drives wheel, so is useful (1)
•        Electrical energy supplies power to drive wheels /
          useful implied (1)
•        Input energy (electrical or energy supplied to motor)
          is converted into heat (1)

         Last point to do with the idea that once moving with constant speed e.g.
•        All work is done against friction
•        No input energy is converted into Ek
•        All input energy ends up as heat
•        Any other relevant point relating to energy (1)                              5

  (b)   (i)      k = F / x
= 1.8 × 106 / 0.90 (1)
= 2.0 × 106 Nm–1 (1)                                                                               2

  (ii)   f = (1 /2π (k/m)0.5 (0)
= (1 /2π (2.0 × 106 / 9.5 × 105)0.5 (1)
= 0.23 Hz (1)                                                                                           2

  (iii)  If wind energy causes this frequency in the structure, the
amplitude increases / resonance occurs / or explanation of
resonance / ref. to natural frequency (1)
e.g. damping is necessary / mass change to shift resonant
frequency / change spring constant (1)                                                    2

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  6.     the pendulum bob is travelling in a circle (1)
so it is accelerating towards the centre (1)
(it has a constant speed in the time interval just before vertical to just
after vertical)

          bob is not in equilibrium (1)
so the tension must be (slightly) larger than the weight of the bob (1)                        3

MAXIMUM 3

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