1. Dynamic forces
Equations of motion

2. 1, Acceleration (rate of change of Velocity) velocity is the vector version of speed, (speed in a straight line)
v – u t
= a
u = initial velocity (in m/s)
v= final velocity (in m/s)
t = time (in seconds (s))
a = m/s2
Can be rearranged to
v = u + at

3. Sample question
A car increases it’s velocity from 10 m/s to 20m/s in 5 seconds. What is it’s acceleration?

4. Sample question (answer)
v – u t
= a
v = 20m/s u =10m/s t = 5 sec
20 – 10 5
= 2m/s2

5. Sample question
A car moving at 10 m/s accelerates at 2m/s2 for 4 seconds. What is it’s final velocity?

6. Sample question (answer)
v = u + at
Can be re-arranged to
u = 10m/s a = 2m/s2 t = 4 sec
v = x 4 = 18m/s

7. Sample question
A stone block with a mass of 800Kg is lifted from rest with a uniform acceleration by a crane such that it reached a velocity of 14m/s. after 10 seconds. Calculate the tension in the lifting cable

8. Sample question To do this calculation we need to use another equation
Force = mass x acceleration
F = ma
Weight (force acting downwards) = mass x acceleration due to gravity
Wt = mg
Wt = m x 9.81 (acceleration due to gravity on earth)

9. Sample question (answer)
First find the acceleration
acceleration
Force due to acceleration
F = ma
Tension
Weight mg
U = 0 from standing start
v – u = a v = a = 1.4m/s2
t t

10. Sample question (answer)
F = ma F = 800 x 1.4 = N
Wt of box mg = 800 x 9.81 = N (7.8kN)
Tension in cable =
= 8938N (8.9kN)
acceleration
Force due to acceleration
F = ma
Tension
Weight mg

11. Force due to acceleration
Sample question
If the stone is lowered at the same rate of acceleration the tension is less than the weight of the box
Force due to acceleration
F = ma
Tension
acceleration
Weight mg
Tension in cable =
= 6728N (6.7kN)

12. Lift
You get the same effect when you travel in a lift. You feel heavier when the lift begins and accelerates upwards and lighter when the lift accelerates downwards

13. A man standing on weighing scales in a lift has a mass of 60 Kg
A man standing on weighing scales in a lift has a mass of 60 Kg. The lift accelerates uniformly at a rate of 2m/s2 Calculate the reading on the scales during the period of acceleration .

14. Force due to acceleration
Weight mg
Force due to acceleration
F = ma

15. F = ma F = 60 x 2 = 120N
Wt of man mg = 60 x 9.81 = N
Force acting on scale =
= 708.6N
The reading on the scale is ÷ 9.81
= 72.23Kg

16. Displacement = average velocity x time
2, Displacement (s) is the vector version of distance (distance in a straight line)
Displacement = average velocity x time
s = (v + u) x t 2

17. Sample question
A car moving at 10 m/s increases it’s velocity to 20m/s in 4 seconds. How far will it have travelled during this time?

18. Sample question (answer)
s = (v + u) x t 2
v =20m/s u = 10m/s t = 4 sec
s = ( ) x 4 2
S = 60m

19. Sample question
How long will it take for an athlete to accelerate from rest to 4 m/s over 8m ?

20. Sample question (answer)
s = (v + u) x t 2
s = v x t 2
u =0 so
t = 2s v
t = = 4 seconds 4

21. 3, Displacement related to acceleration
With zero acceleration (constant velocity)
a = s = ut
s = ut + at2 2
From a standing start ut = 0
s = at2 2

22. Example
A car accelerates uniformly from rest and after 12 seconds has covered 40m. What are its acceleration and its final velocity ?

23. Example
2s = at2
2s = a t2
80 = a
122
80 = a
144
a = 0.56m/s2

24. Finding v a = v – u a = v v = a x t t t v = 0.56 x 12 = 6.7 m/s
From a standing start u = 0
a = v – u t
a = v t
v = a x t
v = x 12 = 6.7 m/s
v = a x t

25. Or alternatively s = (v + u) x t 2 s = v x t 2 v = 80 = 6.7m/s 12
U = 0 from standing start
s = (v + u) x t 2
s = v x t 2
v = 80 = 6.7m/s 12
v = 2s t

26. 4, final velocity, initial velocity acceleration and displacement
v2 = u2 + 2 as

27. Example
If a car travelling at 10m/s accelerates at a constant rate of 2m/s2, what is it’s final velocity after it has travelled 10m?

28. Sample Question (answer)
v2 = u2 + 2 as
V2 = (2 x 2 x 10)
= = 140m/s2

29. D’Alembert’s Principle
When an object is accelerating the applied force making it accelerate has to overcome the inertia. This is the force which resists the acceleration (or deceleration) and is equal and opposite to the applied force. This means that the total force acting on the body is zero
Acceleration (a)
Applied Force (F)
Inertia Force (Fi)
F + Fi = 0 so F = -Fi
F = ma
so ma = - Fi
or Fi = - ma
mass

30. D’Alembert’s Principle
Acceleration
Friction Force
20N
Applied Force
200N
Spring Force
150N
Mass
20Kg

31. Free body diagram F (applied) 200N W (weight) Mg = 20 x 9.81
Fr (Friction)
20N W
(weight)
Mg =
20 x 9.81
=196.2N
F (applied)
200N
Fs (spring)
150N
Free body diagram

32. D’Alembert’s Principle
Downward forces are minus and upward forces are positive.
From the diagram
-200N – 196.2N +20N +150N
= N (which is downward as to be expected)

33. Conservation of energy (Gravitational potential energy)
GPE = mass x gravity x height
(mgh) =
5 x 9.81x10
490.5joules
Work done (in lifting the mass) = force x distance
49.05 x 10
= 490.5joules
(F = mg)
10m
5Kg
Work done = energy gained

34. Kinetic energy
GPE = mgh = joules
Neglecting friction, all the GPE at the top of the slope converts to kinetic energy at the bottom
5Kg
10m
KE (at the bottom) =
490.5J = mv2 2
(v = velocity)

35. The velocity at the bottom v =√(2Ke÷m)
Kinetic energy
Neglecting friction, all the GPE at the top of the slope converts to kinetic energy at the bottom
GPE = mgh = joules
5Kg
The velocity at the bottom v =√(2Ke÷m)
= √ (2 x ÷ 5)
√196.2
= 14m/s
10m

36. Work and energy Work done = energy used distance
Work done (in joules) = Force x distance moved (in direction of force)
Work done = energy used
Force
distance

37. Example
A car of mass 800Kg is stood on a uniform 1 in 10 slope when it’s handbrake is suddenly released and it runs 30 metres to the bottom of the slope against a uniform frictional force of 50N. What is the car’s velocity at the bottom of the slope?

38. Tan of the angle = opposite/adjacent = 1/10= 0.1
Example
First find the angle of the slope
Tan of the angle = opposite/adjacent = 1/10= 0.1
Tan = 5.7o 1 10

39. Example Then find the height of the slope
sin 5.7o = opposite/hypotenuse
Opposite = hypotenuse x sine5.7o
30 x sine 5.7o
=2.98m ht
30m
5.7o

40. Example 1 (conservation of energy method)
Find the gravitational potential energy of the car at the top of the slope
800 x 9.81 x 2.98
= 23.4kN ht
30m
5.7o

41. Example 1 (conservation of energy method)
Work done against friction = force x distance = 50N x 30m = 1.5kj ht
30m
5.7o

42. Example 1 (conservation of energy method)
Kinetic energy of the car at the bottom of the slope
mv2/2 = = 21.9kj
v= √(2 x Ke/m)
v= √(2 x 21900/800)
v= √54.75
v = 7.4 m/s ht
30m
5.7o

43. Example 2 (Resolving forces method)
Work out the forces involved
Weight of car (force acting vertically downward) = mass x gravity
= 800x 9.81
7848N ht
30m
5.7o

44. Example 2 (Resolving forces method)
Work out the forces involved
Example 2 (Resolving forces method)
Wt = weight of car (7848N)
Fn is the normal reaction force of the slope on the car
Fs is the force on the car down the slope
5.7o Fn Fs wt

45. Example 2 (Resolving forces method)
Work out the forces involved
Sine 5.7o = Fs ÷ wt
Fs = Wt x sin 5.7o
Fs = sin 5.7o
Fs = N (Force down the slope)
5.7o Fn Fs wt

46. Example 2 (Resolving forces method)
Work out the forces involved
Resultant force down the slope = Fs – friction force
779.5 – 50
=749.5N
5.7o Fn Fs wt

47. Example 2 (Resolving forces method)
Acceleration down the slope
a= F/m 749.5/800
= 0.94m/s2
5.7o Fn Fs wt

48. Example 2 (Resolving forces method)
Velocity at the bottom of the slope
v2 = u2 + 2as (u2 = 0) so
v2 = 2as
v2 = 2 x 0.94 x 30
v =√56.4
v = 7.5m/s
5.7o Fn Fs wt

49. Linear Momentum and Collisions
Conservation
of Energy 49

50. Momentum = mass x velocity
The total momentum remains the same before and after a collision
Momentum is a vector quantity

51. Linear Momentum and Collisions
A railway coach of mass 25t is moving along a level track 36km/hr when it collides with and couples up to another coach of mass 20t moving in the same direction at 6km/hr. Both of the coaches continue in the same direction after coupling. What is the combined velocity of the two coaches?

52. Linear Momentum and Collisions
Let the mass of the first coach be M1 and the mass of the second coach be M2 and the velocity of the first coach be V1 and the velocity of the second be V2

53. Linear Momentum and Collisions
Before coupling the momentum of the first coach is 25 x 36 = 900tkm/hr and the momentum of the second is 20 x 6 =120 tkm/hr
Which is a total of 1020tkm/hr

54. Linear Momentum and Collisions
After the coupling the momentum of both is the same as before the coupling which is tkm/hr
And the combined mass is 45t

55. Linear Momentum and Collisions
Velocity after coupling is momentum divided by mass
1020÷45
22.6 km/hr

56. Example
A hammer of mass 200Kg falls 5m on to a pile of mass 300Kg and drives it 100mm into the ground
a) Calculate the loss of energy on the impact.
b) Calculate the work done by the resistance of the ground.
c) calculate the average resistance to penetration.

57. Example
Before falling the GPE of the hammer is 200kg x 9.81 x 5m = 9810j
Kinetic energy of the hammer just before impact = 9810j (0.5 x m x v2)
Velocity of the hammer just before impact v2 = (9810)/0.5 x m
(9810)/0.5 X 200 =98.1
v = √98.1 = 9.9m/s

58. thus 500kg x velocity after collision = 1980 kgm/s
Example
Momentum of hammer just before impact = mass x velocity
= 200kg x 9.9m/s = 1980 kgm/s
Momentum before collision = momentum after collision
mass of hammer plus pile after collision = 200kg + 300kg = 500kg
thus 500kg x velocity after collision = 1980 kgm/s
Velocity after collision = 1980 ÷ 500 = 3.96m/s

59. Example
Kinetic energy after collision (0.5 x mass x v2) = x 500 x 3.962
= 3920j
Loss of energy = 9810 – = 5890j
To find deceleration of pile hammer
v2 = u2 +2as (v2 =0)
u2 = - 2as

60. Example
a = - u2/2s
3.962/0.2
- 78.4m/s2
The minus sign means deceleration, Resistive force of the ground = mass x deceleration
500 x78.4
39.2kN

61. Work done by ground = Force x distance
Example
Work done by ground = Force x distance
39.2kN x 0.1 m
3.92kj
(this agrees with the fact that fact that the kinetic energy of the hammer and pile after impact was 3.92kj and zero when the pile stopped moving in the ground