1. WORK , ENERGY & POWER

2. Work is done on an object if an applied force displaces the object in the direction that the force is applied.
DEFINITION
DEFINITION

3. In symbol form: W = F⋅Δx = FΔx cosθ Unit: joule (J) J = N⋅m
applied force (N)
In symbol form:
work (J)
W = F⋅Δx = FΔx cosθ
angle between force and displacement
displacement(m)
Unit: joule (J)
J = N⋅m
Work is a scalar.

4. WORK FA is perpendicular to the displacement: θ = 90° ; cos 90° = 0.
No Work done on an object if the force and displacement are perpendicular to each other.
Consider a man carrying a suitcase with a weight of 20N on a ‘travelator’ moving at a constant velocity.
WORK
FA is perpendicular to the displacement: θ = 90° ; cos 90° = 0.
NO WORK IS DONE and no energy is transferred.

5. W = FΔx cosθ (where cos0° = 1)

6. W = -FΔx (because cos180° = -1)

7. cosθ If force and displacement are in the same
direction:  = 0° (applied, F// slope)
If force and displacement are perpendicular
 = 90° (W = 0 because cos 90° = 0)
If force and displacement are in opposite
direction:  = 180° (friction, F// slope)
(W = - Fx because cos 180° = -1)
cosθ

8. When climbing stairs at a constant speed, work is done by gravitational force.
The muscles must apply an upward force (equal in magnitude but opposite direction to gravitational force) to lift you vertically upwards.
Take note that two forces (applied upward force and downward weight) act on you in such a way that they are balanced.

9. Work, energy and power The resultant/net force is therefore 0 N.
The net work done on you is also 0 J, but the work done by your muscles and the work done by your weight, can be calculated separately.
However, the definition only asks for the applied force.
Work = weight (magnitude only) × vertical displacement
Work, energy and power

10. Examples
A student pushes a lawnmower over a distance of 10,5 m across the lawn. He exerts a force of 190N at a angle of 29º with relation to the horizontal. Calculate the amount of work he does?
A student pushes a lawnmower over a distance of 10,5 m across the lawn. He exerts a force of 190N at a angle of 29º with relation to the horizontal. Calculate the amount of work he does?

11. W = F∆xcos θ
= (190cos29º)(10,5)cos 0
= 1744, 87 J
W = F ∆xcos θ

12. A 220 N force is applied horizontally to a box of mass 50 kg which rests on a rough horizontal surface and the box moves 10 m. The kinetic friction between the surface and the box is 40 N.
Calculate:
1. The work done on the box by the applied force.
2. The work done on the box by the friction.
3. The net work done on the box.
4. The net force acting on the box.

13. 1. Wa = Fa∆xcos θ
= 220(10)cos 0
= 2200 J
2. Wf = Ff∆xcos θ
= 40(10)cos 180
= -400 J

14. Wnet = Fnet∆xcos θ
=(Fa+(-ff) ∆xcos θ
=(220-40)(10)cos 0
= 1800 J
4. Wnet = Fnet∆xcos θ
1800 = Fnet(10)cos 0
Fnet = 180N

15. A crate of mass 70 kg slides down a rough incline that makes an angle of 20° with the horizontal, as shown in the diagram below. The crate experiences a constant frictional force of magnitude 190 N during its motion down the incline. The forces acting on the crate are represented by R, S and T.

16. 1. Label the forces R, S and T.
2. The crate passes point A at a
speed of 2 m·s–1 and moves a
distance of 12 m before reaching
point B lower down on the incline.
Calculate the net work done on the crate during its motion from point A to point B.

17. Wnett = Fnett ∆x cos ∂
=( F// + Ff) ∆x cos ∂
= ((70)(9,8) sin 20˚ + (-190))(12)cos 0˚
= 535, 51 J √

18. YOU TRY
Calculate the net work done on a trolley where a force of 30 N is applied to the trolley. The trolley moves 3 m to the left. The force of friction is 5 N to the right.

19. NET WORK ON AN OBJECT Work done by normal force:
Work done by frictional force:
Work done by gravity

20. The work-energy principle
The work done on an object by the net force is equal to the change in kinetic energy of the object.
Symbol form: Wnet = ΔEK
= EKf - EKi

21. Fnet⋅Δx cosθ = ½mvf2 - ½mvi2
change in kinetic energy
(J)
Displacement (m)
Wnet = ΔEK
Fnet⋅Δx cosθ = ½mvf2 - ½mvi2
initial velocity (m⋅s-1)
mass
(kg)
Displacement (m)

22. Examples
The diagram shows a crate of mass 90 kg sliding down a rough surface inclined at 20° to the horizontal. A constant force F, parallel to the incline, is applied to the crate. The crate moves down the incline at constant velocity
20°
20°

23. The magnitude of the kinetic frictional force (fk) between the crate and the surface of the inclined plane is 266 N.
a. Draw a labelled free-body
diagram showing all the forces
acting on the crate.

24. What is the magnitude of the
net work done on the crate?

25. c. Calculate the magnitude of the force F.

26. The crate is released from rest without force F being applied, it moves 3 m down the incline.
d. Using energy principles only, calculate
the speed of the block after the 3 m.

27. A skateboarder is practising a sequence of tricks at the local skate park on a half- pipe. The total mass of the skateboarder and skateboard is 75 kg. The skater leaves point A, 2,4 m above the ground. He skates down the ramp towards point B. He reaches Point B, 1,6 m above the ground, with a speed of 3,75 m·s-1 just by rolling along and without using his feet to push himself along the half-pipe.
The skateboarder has not oiled the wheels of his skateboard for some time, so there is significant friction between the axles and the wheels of the skateboard

28. State in words, the work-energy
theorem.
b. Calculate the work done by the
gravitational force on the
skateboarder as he moves from
point A to Point B.
c Using the work energy theorem,
determine the work done by the
frictional force exerted on the
skateboard.

29. A wooden block of mass 2 kg is released from rest at point P and slides down a curved slope from a vertical height of 2 m, as shown in the diagram below. It reaches its lowest position, point Q, at a speed of 5 m·s-1.

30. Use the work-energy
theorem to calculate the
work done by the average
frictional force on the
wooden block when it
reaches point Q.

31. The wooden block collides with a stationary crate of mass 9 kg at point Q. After the collision, the crate moves to the right at 1 m·s-1.
Calculate the magnitude of the velocity of the wooden block immediately after the collision.
The total kinetic energy of the system before the collision is 25 J. Use a calculation to show that the collision between the wooden block and the crate is inelastic

32. Forces Conservative forces gravitational force - conservative force
Work done is independent of the path taken.
The net work done in moving an object in a closed path which starts and ends at the same point is zero.
Forces
gravitational force - conservative force
The work done by gravity on each ball is independent of the path
taken

33. Forces Non-conservative forces
The work done is dependent of the path taken.
The net work done in moving an object in a closed path which starts and ends at the same point is not zero.
Forces
Examples:
friction and air resistance
Applied force
The work done by FA is more when the longer path is taken.
Force of gravity and the F// slope not n/c force

34. Definition conservative/ non-conservative force
A force for which the work done in moving an object between two points is:
Non-conservative = dependent on path taken
Conservative = independent on path taken

35. Wnc = ΔEK + ΔEP = (½ mvf 2− mvi2)+ (mgh-mgh)
Non-conservative forces
Wnc = ΔEK + ΔEP = (½ mvf 2− mvi2)+ (mgh-mgh)

36. EXAMPLE:
An 800 kg car traveling at 15 m·s−1 down a 30° hill needs to stop within 100 m to avoid an accident. Using energy calculations only, determine the average force that must be applied to the brakes over the 100 m.

37. YOU TRY
A worker applies a constant force of 45 N on a crate of mass 25 kg, at an angle of 30° with the horizontal. When the crate reaches point P, its velocity is 12 m.s-1 and 3,5 m further it reaches point Q at a velocity of 10,8 m.s-1

38. 1. Draw a labelled free-body diagram to show the horizontal
forces acting on the crate during its motion. The length of the
vectors should be an indication of their relative magnitudes.
2. Write down the NAME of the non-conservative force that
opposes the forward motion of the crate.
3. Use ENERGY PRINCIPLES to calculate the magnitude of the
non-conservative force mentioned in QUESTION 3.2.

39. Work/Force with friction
Wnet = Fnet x cos
Wnc = ΔEk + ΔEp
Wnet = Ek (work energy theorem)
The work done by the resultant (net) force acting on an object is equal to the change in the kinetic energy of the object.
If a = 0 (constant v or v = 0), then Wnet = 0 (Fnet = 0 or Ek = 0)
Work with friction

40. Energy Potential Energy (EP): The energy an object has
due to its position
Gravitational EP: the energy of an object due to it’s
position above the earth (EP = mgh).
Kinetic Energy (EK): The energy of an object in
motion (EK = ½mv2)

41. Work/Force without friction
Conservation of Mechanical Energy
Vertical projectile motion
Pendulum (Tarzan swing)
EM at A = EM at B
(EP + EK)A = (EP + EK)B
(mgh + ½ mv2)A = (mgh + ½ mv2)B
Mechanical energy in an isolated system is conserved provided the sum of kinetic and gravitational potential energy remains constant

42. Law of Conservation of Mechanical Energy
EXAMPLE : Pendulum The 2 kg pendulum swings from A at 5 m.s−1 to B, on the ground, where its velocity is 8 m.s−1. Determine the height at A.
(Ep+Ek)A = (Ep+Ek)B
(mgh + 0.5mv2)A = (mgh + 0.5mv2)B
(2)(9,8)h + 0.5(2)(52 ) = (2)(9,8)(0) + 0.5(2)(82 )
19,6h + 25 =
h = 1,99 m

43. A roller coaster car starts at rest at point A. The
track from point A to point B has no friction
while the track from point B to point C is rough.
The mass of the roller coaster car with
passengers is 1000 kg.

44. 1 Calculate the speed of the car at point B.
2 If the car arrives at point C with a speed of 10 m·s-1,
calculate the amount of work done by the frictional
force while it is moving from B to C.
3 If the frictional force of the whole track is taken into
account and the car moves from A to C over a
distance of 200m, find the average frictional force,
(assume that it remains constant), on the car. The
car comes to a standstill at point C.
7. The arch over

45. P = Power (W)
W = Work (J)
E = Energy (J)
t = Time (s)
Power is the rate at which work is done OR the rate at which energy is expended.

46. AVERAGE POWER (constant v)
We can calculate the average power needed to keep an object moving at constant speed using the equation:
Paverage = Fvaverage

47. A man lifts a 50 kg bag of cement from ground level up to a height of 4 m above ground level in such a way that the bag of cement moves at a constant velocity (i.e. no work is done to change kinetic energy). Determine his average power if he does this in 10 s.