Power

Work, Energy and Power

(Use g = 9.8 Nkg^-1)

Examples

A Find the work done in lifting a mass of 6 kg through a height of 10 m.

W = Fx W work done in J

F force in N

x distance moved in the direction of the force in m

W = 6 x 9.8 x 10

= 588 J

B) An escalator carries 100 people of average mass 70 kg to a height of 6 m in one minute.

Find the power necessary to do this.

Work done = F x = 100 x (70 x 9.8) x 6 J

Power = rate of doing work = work done per second

= (100 x 70 x 9.8 x 6) / 60 s = 6860 W or 6.86 kW

1. Find the work done to lift a mass of 10 kg through a height of 5 m. If this takes 15s what is the power?

W= Fx = (10kg)(9.8Nkg^-1) (5m) = 490J

P = W/t = 490 J / 15s = 32.67 W

2         A mass of 5 kg is pulled along a surface at a constant speed. Frictional force between the mass and the surface is 4N what work is done in moving the mass 2 m along the surface? Find the power if this takes 2 s.

W = Fx = (4N) (2m) = 8J

P = W/t =  8J/ 2s = 4 W

3        A girl of mass 50 kg runs up a flight of steps 4.5 m high in 5 s. What power does she develop?

W = Fx             F = 50kg x 9.8 = 490.5 N                     x = 4.5m

W = 490.5 x 4.5 = 2207.25 J

P = W/t = 2207.25 J / 5s = 442w

4        2 x 10⁶ kg of water per second flow over a large waterfall of height 50 m.

W = Fx                   F = 2 x 10⁶ kg x 9.8 = 1.96 x10⁺⁷N      x=50m

W= (1.96 x10⁺⁷N) (50m) = 9.8 x10⁺⁸ J

What power is available?

As 1W = I J per second the water falling over the fall transfers 9.8 x10⁺⁸ J of GPE per second the available power is 9.8 x10⁺⁸ W (ignoring losses due to friction)

5        The first-stage rocket motors of Saturn 5 burnt fuel at the rate of 13600 kg per second. The work done by the pumps to drive the fuel into the combustion chambers is the same as that necessary to lift the fuel through 1680 m.

W= Fx                    F = 13600kg x 9.8= 1.3328 x 10⁺⁵ N    x = 1680m

W = (1.3328x 10⁺⁵ ) (1680) = 2.24 x 10⁺⁸ J

What power is needed to do this? (Your answer is about twice the engine power of the largest ocean liner.)

As this mass of fuel is burnt in 1 second then 2.24 x 10⁺⁸ J of work is done in 1 second so the power is 2.24 x 10⁺⁸ W

6        A gas turbine locomotive developing 6330 kW pulls a train at a steady velocity of 108 kmh^{- 1}. What is the total force in Newtons resisting the motion?

At a steady velocity all work done by the engine is against friction or gravity.

The engine does 6.33 x10⁶ J of work every second

As W = Fx the force exerted by the engine is equal Work done per second divided by the distance travelled per second

F = W/x            108 kmh^-1 = (108000m)/(60 min x 60 sec) = 30 ms^-1

F = 6.33 x10⁶ J/ 30m = 211 000N = 2.11 x 10⁵ N

7        A mechanical loader driven by an engine developing 3 kW lifts a total load of 200000 kg to a height of 3 m in one hour. What is the efficiency of the system? 54.4%

Work done lifting 2 x 10⁵kg through 3m

W= Fx              F = (2 x 10⁵kg) (9.8) = 1.96 x 10⁶ N     x = 3m

W = (1.96 x 10⁶ N) (3m) = 5.88 x 10⁷ J

Time = 1hr = 3600 seconds

Power out = (5.88 x 10⁷ J) / (3600) = 1630 W

(Efficiency = Power out / Power in = 1630 / 3000 = 0.54 = 54%)

Questions

For each question show all your working clearly.

1.      A lot of energy is wasted in a car. For every
100 J of chemical energy in the petrol, only
25 J are transferred to useful kinetic energy.
The rest just heats up the engine and the air.

a)      Calculate the efficiency. 25%

2.       The diagram shows the energy transfers for a
Bunsen burner heating a beaker of water.
What is its efficiency as a water heater? 40%

3.       In a solar cell, for every 80 J of solar energy
shining on it, only 4 J is transferred to useful
energy (as electricity).

a)      What happens to the other 76 J? reflected, lost as low grade hear

b)      What is its efficiency? 5%

c)       

4.       A pulley system lifts a load and gives it
6000 J of potential energy. The person
pulling on the rope gives it 8000 J of energy.
What is the efficiency?  75%

5.       An electric kettle has a power rating of 2 kW
and is switched on for 100 seconds.
While heating up, it loses 60 000 J to the
surroundings.

a)      How much energy is supplied to the kettle?
(1 kW = 1000 W = 1000 joules per second)200kJ

b)      How much is given to the water? 60kJ

c)      What is the efficiency of heating water? 70%

6.       An electric motor on a building site has a
power rating of 400 W and lifts a load of
bricks weighing 600 N through a height of
10 m in 20 seconds.

a)      How much energy is needed to lift the bricks? 6000J

b)      How much energy is supplied to the motor
in 20 seconds? 8000J

c)      What is the efficiency of the motor in
doing this job? 75%

Efficiency

1. An energy efficient light bulb is rated at 20W. It produces 5W of light. Calculate its efficiency.

5/20 x 100 = 25%

2. Christine 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 32/40 x 100 = 80% (fallen into the trap!)

3. Geraint 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 Geraint’s bike. 1577.6/1600 x100 = 98.6%

4. The coal in Thomas’ boiler contains 40 kJ of energy. He does 2576J of useful work as he puffs along a branch line. Calculate Thomas’ efficiency.

2.576/40 x100 = 6.44%

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
6. 20.1/60 x100 = 33.5%

7. 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 calculate its efficiency. 2.685/3 x100 = 88.6%

8. 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. Calculate the maximum wattage of the chemical energy in the fuel burnt. 700 bhp x 750W = 5.25 x 10⁵ W

5.25 x 10⁵ x 100 /output = 30%

5.25 x 10⁷ / output = 30

Output = 5.25 x 10⁷ / 30 = 1.75 x 10⁶  W

Of the 70% of energy is that is not usefully converted, 55% heats the cooling water that surrounds the engine block 0.55 x 1.75 x10⁶ W = 9.625 x 10⁵ W whilst  15% may be in the hot exhaust gases. 2.625 x 10⁵ W. Calculate the power lost by the radiator and the exhaust.

To make car engines more efficient the fuel has to burn at a higher temperature and the exhaust must be kept cooler. Where should Jeremy move to improve the efficiency of his car? Pluto

Forces in action assessed homework

 

3.2 Forces in action MS

1)      1) A

2)      2) B

3)      3)D D

4)     4)  F = m x a = 24.1 x 3.5 [1]

F = 84.4N [1]

5)     5  a = F/m = 18/0.61 [1]

a = 30ms^-2 [1]

6)      6 F = m x a = 60 x 2.3 = 138N [1]

Therefore force on 2^nd skater = 138N [1]

a = F/m = 138/55 = 2.5ms^-2 [1]

7

8)Moment = Fd = 73.1 x 0.25 [1]

Moment = 18.3Nm [1]

 

9) Anticlockwise = 450 x 1.5 = 675Nm [1]

Clockwise moment = 675 = 500d; d = 675/500 = 1.35m [1]

 

10) . (i)      1        3600 ´ 1.0 = X ´ 2.5                                                                                      C2

one mark for one correct moment, one mark for the

second correct moment and equated to first moment       

                                                                                                                               A0

2        X = 1440 (N)                                                                                                  C1

Y = 3600 – 1440        or 3600 ´ 1.5 = Y ´ 2.5                                                A1

    = 2160 (N)                                                                                                  B1

(ii)     Not a couple as forces are not equal                                                              B1

and not in opposite directions / the forces are in the same direction

y12 Homework on GPE and KE

Gravitational Potential Energy & Kinetic Energy

acceleration due to gravity = 9.81 ms^-2

1.       A car has a mass of 750 kg. Calculate its kinetic energy at the following velocities

          (a) 10 ms^-1  (b) 15 ms^-1  (c) 20 ms^-1  (d) 30 ms^-1  (e) 35 ms^-1  {NB 1 ms^-1  = 2 m p h}

v (ms-1)	ke
10	3.750 X 10 +04	J
15	8.438 X 10 +04	J
20	1.500 X 10 +05	J
30	3.375 X 10 +05	J
35	4.594 X 10 +05	J

2.         (a) A car has a mass of 1000 kg. Calculate its KE at a velocity of 25 ms^-1  .

 3.125 x 10⁵ J

          (b) A train has a mass of 37 tonnes {1 tonne = 1000 kg}. If it also has a velocity of 25        ms^-1  what is its  kinetic energy ?

1.156 x 10⁷ J

3.       A woman has a mass of 65 kg.What is her GPE at the top of a 12 metre diving board ?

7651.8 J

to 3 sig figs 7650 J

4.         A cat has a mass of 6 kg. What is its GPE at the top of a tree that is 4.2 m above the        ground ?            247.212 J

5.       The Eiffel Tower is 300 m high. What is the GPE of a 100 g bird perched on the top of it ?

            ( = 01Kg)  294.3 J

 

6.       The Great Pyramid of Khufu (Cheops to the Greeks) is 146 m high. It is made of stones of     mass 250 t. What is the GPE of the topmost stone ?

            3.58 x 10⁸J    (m = 2.5 x 10⁵ kg)

7.       The Empire State Building is 449 m high(but this includes a 68 m TV mast).What is the    GPE of a ball of mass 400 g at the top of the building (not the TV mast) ? If the ball is   dropped over the side what will happen to the GPE ? What will be its speed at the instant           before it hits the ground ? (Ignoring air resistance)

            GPE = 1495 J

            Ignoring air resistance GPE= KE             mgh=1/2 mv²           m cancels v² = 2gh

            So v = √2gh = √2 x 9.81(449 -68)      = 86.5 ms^-1

8.       Olympus Mons is a volcano on Mars. It is 25 km high. What would be the GPE of a          human of mass 50 kg on its summit ? (surface gravity = 0.38 that of Earth) Compare this    with the same human standing on the summit of Everest 8534 m high.

            Olympus mons GPE = mgh = 50 (9.81 x 0.38) 25 000m = 4.66 x 10⁶ J

Mt Everest = 4.19 x 10⁶ J (Mars is much smaller than the Earth, a 30Km high mountain would collapse in the Earth’s gravitational field.

9.       The space shuttle has a mass of 80 t. Escape velocity (the velocity an object has to travel     at in order to escape the earth's gravitational field) is 11 km/s. What is its KE at this     speed? If all this KE is turned into GPE how high would it be ? What assumption do you    have to make in your calculation. 

            KE = 4.84 x 10¹² J = GPE = mgh so h = KE/mg = 6.16 x 10⁶ m

Assuming g is constant at 9.81. g decreases as a function of 1/r². r = distance from the centre of the earth. When this is taken into account the height is ∞

How much energy would it take to accelerate the space shuttle to light speed (300 000 kms^-1  ). {This is not possible, also the equation you will use to calculate KE is not valid near the speed of light but it will do for this purpose}

3.6 x10²¹ J

A nuclear power station produces around 500 MW

(500 000 000 J per second). How long would 100 power stations take to produce this amount of energy?

100 power stations 5 x 10⁹ Js^-1

Time to produce energy to accelerate space shuttle = 3.6 x10²¹ J / 5 x 10⁹ Js^-1 = 7.2 x 10¹¹ s

22832 years (NB same amout of energy required to slow down when you get there)

How feasible is travelling close to the speed of light?

The nearest star is about 4 light years away. How long would it take the space shuttle to get there at 11 kms^-1  (22 000 mph)? {1 light year is the distance light will travel in 1 year}

How feasible is interstellar travel?

Distance = 4 (3x10⁸ ms^-1) (365d x24 hr x 60 min x 60s) = 3.78432x 10¹⁶ m

Time = distance/ velocity = 3.78432x 10¹⁶ m / 11 000 ms^-1 = 3.44029x10¹² s = 109091 years

About 10 times longer than human civilisation!

Homework on Work

Work

Take g as 9.8 ms-2 or 9.8 Nkg-1

How much work is done if you push a shopping trolley with a constant force of 60N and it moves 5m in a direction parallel to the force?

W = Fx

W = (60N) (5m)

W  = 300 J

A delivery driver lifts a mass of 6.5kg onto the back of a lorry 1.5m from the ground. How much work is done in this energy transfer?

W = Fx                        F = (6.5kg) (9.81 Nkg^-1)                    x = 1.5 m

W  = (6.5kg) (9.81 Nkg^-1) (1.5 m)

W = 95.55 J

How much work is done lifting a 5kg bag 1.2m and place it on the table. In your calculation you assume all the work is done against what?

W = Fx                        F= (5kg) (9.81 Nkg^-1)                        x = 1.2 m
W = (5kg) (9.81 Nkg^-1) (1.2m)

W = 58.8 J

Work is done against gravity

How much work is done pulling a bag of rubbish 10m across a field by pulling on a string at 40⁰ to the horizontal with a force of 250N?

W = F cos θ x

W = (250 N) (cos 40) (10 m)

W = 1915 J

A toy car has a mass of 110g and its clockwork engine exerts a force of 0.12N. Unfortunately it is not well made and its wheels are at a 13⁰ angle to its direction of motion. What work does it do in travelling 25cm along a heavily carpeted floor?

W = F cos θ x                        x = 25cm = 0.25m

W = (0.12N) (cos 13) (0.25 m)

W = 0.029J

A pyramid builder is organising his gang of acolytes to pull a large stone block up a ramp. The stone weighs 2.5t and the ramp has a height of 7m. The ramp is 150m long and the acolytes exert a force of 1.2 kN. How much work to they do? How much work is done on the stone to lift it through a height of 7m? How much work is done against friction?

Work done lifting stone = weight of stone  x vertical height

W = Fx                        F = (2.5 x 10³ kg) (9.81 Nkg^-1)         x = 7m

W = (2.5 x 10³ kg) (9.81 Nkg^-1) (7m)

W = 1.72 x 10⁵ J

Work along ramp = Tension in rope x length of slope

W= Fx                         F= 1.25 kN = 1.2 x 10³ N                  x =150m

W = (1.25 x 10³ N) (150m)

W = 1.88 x 10⁵ J

Work done against friction= work done pulling stone up ramp – work done against gravity

W against friction = (1.88 x 10⁵ J ) – (1.72 x 10⁵ J) = 16000 J

An 80kg baseball player slides to a halt from a speed of 8 ms^-1 in a distance of 4m. What is the average stopping force exerted on him by the ground? How much work is done on him? Where does the energy come from?

We need to calculate the force exerted on the baseball player by the ground. For this we need to know his acceleration.

Using                          v² = u² + 2as 

Rearranging  2as = v² – u²

Cross multiplying       a = (v² – u²) / 2s         v= 0     u = 8 ms^-1 s = 4m

 a = (0 - 8²) / 8m = 64/8 = 8ms^-2

Using Newton’s second law F=ma = (80 kg) (8ms^-2) = 640 N

W= Fx             F= 640N                     x = 4m

W = (640 N) (4m)

W = 2560 J

KE of player

A 50g ball bearing is dropped from a height of 50cm into a tray of fine sand and embeds itself to a depth of 1.5cm. What was the average vertical force exerted on the ball bearing by the sand? How much work is done on the ball bearing?

Ignoring frictional losses

Work done on sand = Work done lifting ball to height of 50cm

Work done lifting ball = Fx                F = (0.05kg) (9.81 Nkg^-1)     x = 0.5m

W = Fx = (0.05kg) (9.81 Nkg^-1) (0.5m) = 0.25 J

Work done on sand = average force exerted by sand x depth of ball

            W= Fx

F= W/x = 0.25J / 0.015m = 16.4 N

Assessed Homework Electromagnetism

 

1.       (a)     (i)      equally spaced horizontal parallel lines from plate to plate (1)
arrows towards cathode (1)                                                                     2

  (ii)   ½ mv2 = qV; v = √(2eV/m) = √(2 × 1.6 × 10–19 × 7000/9.1 × 10–31) so (1)
v = 4.96 × 107 (m s–1) (1)                                                                        2

  (b)   (i)      arrow perpendicular to path towards centre of arc (1)                             1

(ii)     out of paper/upwards;using Fleming’s LH rule (for conventional
current) (2)                                                                                              2

  (iii)  mv2/r; = Bqv; r = mv/Bq ;= 9.4 × 10–2
(m)                                                                                                          4

  (c)   change magnitude of current in coils to change field; (1)
change field to change deflection; (1)
reverse field/current to change deflection from up to down (1) max 2             2

[13]

 

  2.     At least 3 field lines inside solenoid parallel to axis; (1)
Lines equally spaced over some of length of solenoid. (1)
Arrows on lines pointing left to right. (1)                                                                   3

[3]

 

 

 4.      (i)      I = V/R = 12/50 (1)
= 0.24 A (1)                                                                                                                2

  (ii)   Power in primary = power in secondary / IpVp = IsVs (1)
Ip = 0.24 × 12 / 230 = 0.0125 A (1)                                                                  2

[4]

 

 

5.      

  (b)   (i)      k.e. = QV; = 300 × 1.6 × 10–19 = (4.8 × 10–17 J) (2)                                2

  (ii)   1/2mv2 = 4.8 × 10–17; = 0.5 × 2.3 × 10–26 × v2 so v2 = 4.17 × 109;
(giving v = 6.46 × 104 m s–1) (2)                                                             2

                   (d)     (i)         semicircle to right of hole (1) ecf(a); (a) and d(i) to be consistent        1

(ii)     mv2/r; = BQv; (2)
giving r = mv/BQ = 2.3 × 10–26 × 6.5 × 104/(0.17 × 1.6 × 10–19); (1)
r = 55 mm;so distance = 2r = 0.11 m (2)                                                 5

[13]

 

Bridges prep

 

Moments prep