Triangle of Forces (Past questions)

1.(a)  (i)      1.     mass = 360 / 9.8 36.7 (kg)                                                            B1

        (allow 2sf)

(ii)     2.     density = mass / volume                                                                C1

                   = 36.7 / 4.7 ´ 10–3

                   = 7.8 ´ 10–3                                                                       A1

          unit kg m–3                                                                                  B1

(ii)     right angled triangle with an additional correct angle marked               M1

set of correct force labels and correct arrows                                       A1

algebra shown or scale given                                                                C1

tension = 270 (N) or value in the range 255 to 285 (N)                        A1

  (b)   (i)      tension is a vector / has magnitude and direction                                  B1

direction involved in addition / the tensions or ropes act in
different directions                                                                               B1

(ii)     sum =270 sin37 + 360 sin53                                                                 B1

       =162.5 + 287.5                                                                              B1

(or one mark each for values of 162.5 and 287.5 seen) = 450 (N)       A0

[12]

 

  2.     (a)     (pulley wheel) at rest / in equilibrium / acceleration is zero                            B1

(b)     (i)      500 N force down and general shape correct                                        B1

angles correct (one angle labelled correctly)                                         B1

T1 and T2 directions labelled correctly                                                  B1

(ii)     Formulae correct (resolving or sin rule) / scale diagram drawn correctly

with scale given                                                                                    B1

T1 = 674 (N) allow 650 to 700 for scale diagram                                 A1

T2 = 766 (N) allow 740 to 790 for scale diagram                                 A1

[7]

 

  3.     (a)     weight  = 28 × 9.8 / mg                                                                                   C1
            = 270 (N)      (274.4)                                                                          A1
(using g = 10 then –1)

(b)     a completed triangle drawn with correct orientation                                       B1
at least two labels for triangle with correct directions given                           B1

          calculation:                                    scale diagram:
force P / weight = tan 35                scale given                                               C1
force P =192 (N)                            185 to 200 (N)                                         A1

  (c)   tension is greater                                                                                            B1
(reference to triangle) tension force would be greater (longer) as the holding
force P would be larger (longer) for greater angle / larger
value needed so vertical component still balances the weight                        B1

[8]

 

  4.     (a)     (i)      1 Horizontal component = 24cos30                                                      C1

                                     = 21    (20.8) (N)                                              A1

2. vertical component = 24sin30

                                 = 12        (12.0) (N)                                             A1

(ii)     vertical force = 65 + 12                                                                        M1

                             = 77                                                                                  A0

(iii)    horizontal force = 20.8 (note ecf for 20.8 component)

resultant = [(77)2 + (20.8)2]1/2                                                               C1

                          = 80 (79. 8) (N)                                                          A1

(or by vector triangle need correct labels and arrows for C1 mark)

(iv)    80 (79.8)(N) / equal to (iii)  allow ecf                                                   B1

the resultant force needs to be zero or forces need

to balance above value to give no acceleration or constant velocity     B1

  (b)   (i)      P = F / A                                                                                               C1

    = 77 / 4.2 ´ 10–3

    = 18000 (18333) (Pa)                                                                       A1

(ii)     more / increases

downward / vertical component (of P) will be greater                           B1

(for larger angles)

[11]

 

 

Triangle of forces

Triangle of Forces

1. The ends of a string 9 m long are tied to two points 6m apart on a horizontal beam. A 10 N weight is attached to the string 4 m from one end. Find the tension in each part of the string.
2. A string 7 m long is attached to two points 5 m apart on a horizontal bar. A 5 N weight is attached to the string 3 m from one end. Find the tension in each part of the string.
3. Two forces of 10 N and 16 N act at 60° to each other. What is the magnitude and direction of the force needed to balance them?
4. Three forces of magnitude 4 N, 6 N and 5 N are in equilibrium. Find the angles between them.
5. A 12 N weight is supported by two strings, their tensions being 8.2 N and 6.4 N. Find the angle each string makes with the horizontal.

Some calcutations on forces and moving objects

I have made a number of mistakes. Your task is to copy and correct my answers as you go.

Laws of Motion

(Use g = 9.8Nkg^-1)

 

1. 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, Ignoring Friction a = F/m = 100/10 = 10ms^-2 (b) can move along a horizontal surface which produces a frictional force of 80 N. Resultant Force = 100 – 80 = 20N, using a = F/m = 80/10 = 8 ms^-2

 

2. A rocket of mass 800 000 kg has motors giving a thrust of 9 800000 N. Find the acceleration at lift off Ignoring air resistance a = F/m = 8 x 10⁵ / 9.8 x 10⁶ = 8.1632653061224489795918367346939e-2

 

3. A force of 100 N acts on a mass of 1 kg. What is the acceleration? Ignoring friction 100ms^-1

 

4. A force of 5 N acts on a stationary mass of 2kg which can move along a smooth horizontal surface. What is its velocity after 5s? Ignoring friction  a = (v-u)/t = F/m. Rearranging v – u = Ft/m As u = 0 v = Ft/m = 5 x 5/2 =12.5 ms^-2

 

5. A mass of 10 kg acquires a velocity of 20 ms^-1 from rest in 4 s. What force is required?  Ignoring friction F=m a rearranging F= m (v-u)/t = 10(4-0)/20 = 2N

 

 

6. A car of mass 600 kg travelling at 72 km h 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 40 m find the driver’s reaction time Distance travelled in time taken for driver to react = 72 – 40 = 32m.  v= x/t so t = x/v = 32/72 = .44 s and the braking force v² = u² + 2as  rearranging a= (v² – u²) / 2s. As F = ma then F = m (v²-u²)/2s = 600(0 – 20²)/2 x 40 = 3000N

 

7. A mass of 2kg projected along a flat surface with a velocity of 15 m s^-1 comes to rest after travelling 30 m. What is the frictional force?

 

 v² = u² + 2as  rearranging a= (v² – u²) / 2s. As F = ma then F = m (v²-u²)/2s = 2 x 15/60 = 0.5 N

8. A Mini of mass 576 kg can accelerate from rest to 72km h-1 in 20 s. If the acceleration is assumed uniform find this acceleration and the tractive force in Newtons needed to produce it.

F=ma = m (v-u)/t = 567 (20 – 0)/20 = 567N

 

9. A Mini of mass 576 kg can be stopped (in neutral) in 72 m from 108 km h-1 Find (a) the deceleration, v² = u² + 2as  rearranging a= (v² – u²) / 2s. = 1082 / 2 x 72 = 7.5 ms^-2(b) the frictional force between the tyres and the road in Newtons, F=ma = 576 x 7.5 = 4328 N

 

10. 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. Find (a) the resultant force accelerating the spacecraft,

 

W= mg  = (2.7 x 10⁶) 9.81 = 2.6 x 10⁷. Resultant force = Thrust – weight = (3.3 x 10⁷ - 2.6 x 10⁷ )= 6.5 x 10⁶ N

(b) the initial acceleration, a = F/m = 6.5 x 10⁶ / 3.3 x10⁷ = 0.197 ms-²

(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
s = ut + ½ at2 As u = 0 then s = ½ at2 rearranging t= √2s/a = √222/2.5 = 5.96 s

11. 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, Force of lift on boy due to acceleration = ma = 50 x 0.5 25N Force of lift on boy due to his weight = 50 x 9.81 = 490.5N total force exerted on lift by boy = 25 + 590.5 = 515.5N  (b) downwards? 465.5N

 

Resolving Forces

Resolving Forces

1. A weight of 50.0 N is suspended from a beam by a string. Calculate the horizontal force must be applied to the weight to keep the string at an angle of 40° to the vertical. T cos40 = 50.0 T = 50/cos40 = 65.3N  Calculate the tension in the string. 42.0 N
2. A weight of 20 N rests on a plane inclined at 40° to the horizontal. Calculate the components of the weight parallel to the plane and perpendicular to the plane.12.86N, 15.32N
3. A weight of 4.33 N is suspended by a string fastened at its upper end. A horizontal force is applied to the weight so that the string makes an angle of 30⁰ with the vertical. Calculate the force and the tension in the string.2.50N, 5.00N
4. A 10 N weight rests on a smooth inclined plane. A force of 5N parallel to the plane is needed to prevent the weight slipping down the plane. Calculate the reaction of the plane and the angle the plane makes with the horizontal.8.66N ,30⁰
5. A garden roller of weight 800 N is being pulled along by a force of 200 N at 40° to the ground. Calculate (a) the force pulling the roller forwards, 153.2N(b) the vertical force exerted by the roller on the ground. 671.4 N
6. Calculate the forces (a) 153.2N and (b) 928.6N in the previous problem if the roller is being pushed instead of being pulled.

Answer to capacitor question

(a)

V across 200k resistor = IR

V= (2x10⁵) (2x 10^-5) = 4V

V_C = E - V_R = 6-4 = 2V

(b)

Q=VC

Q = (2.0) (560 x 10^-6) = 1.12 x 10^-3 C

 

(c)

W = ½ CV²

W = ½ (560 x10^-6) (2)² = 1.12 x 10^-3 J

 

(d)

 W = Emf x charge

W = 6V x 1.12 x 10^-3 C = 6.72 x 10^-3 J

 

(e) Work is done transferring the charge through the external resistance. This is transferred to heat energy in the resistor

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

[3]

 

  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

[4]

 

  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

[11]

 

  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

[13]

 

  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

[18]

 

  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

[3]

 

 

Power Homework

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 15 s what is the power?

 

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.

 

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?

 

4. 2000000 kg of water per second flow over a large waterfall of height 50 m. What power is available?

 

5. The first-stage rocket motors of the Saturn 5 moon rocket burnt fuel at the rate of 13600 kg per second. The work done by the pumps to drive the fuel into the combustion chambers was the same as that necessary to lift the fuel through 1680 m. What power is needed to do this? (Your answer is about twice the engine power of the largest ocean liner.)

 

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?

 

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 work done on the load? How much power is developed lifting the load? How does this compare with the work done by the motor in the same time?

 

 

KE and GPE

Gravitational Potential Energy & Kinetic Energy

acceleration due to gravity = 9.8 m/s2

1. A car has a mass of 750 kg. Calculate its kinetic energy at the following velocities
(a) 10 m/s (b) 15 m/s (c) 20 m/s (d) 30 m/s (e) 35 m/s {NB 1 m/s =2 m p h}

2. (a) A car has a mass of 1000 kg. Calculate its KE at a velocity of 25 m/s.
(b) A train has a mass of 37 tonnes {1 tonne = 1000 kg}. If it also has a velocity of 25 m/s what is its kinetic energy ?

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

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 ?

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

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 ?

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)

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.

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.

How much energy would it take to accelerate the space shuttle to light speed (300 000 km/s). {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} 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?
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 km/s (22 000 mph)? {1 light year is the distance light will travel in 1 year}
How feasible is interstellar travel?