1. WORK, POWER & ENERGY part 1

2. Work is the product of the component of the force exerted on an object in the direction of the displacement and the magnitude of the displacement.
W = FΔd

3. WORK
In order for work to be done, three things are necessary:
There must be an applied force.
The force must act through a certain distance, called the displacement.
The force must have a component along the displacement.

4. A teacher applies a force to a wall and becomes exhausted.
Read the following statements and determine whether or not they represent examples of work.
A teacher applies a force to a wall and becomes exhausted.
NO, displacement doesn’t occur
A book falls off a table and free falls to the ground.
Yes, displacement in the direction of force

5. A truck carries a box in it’s bed 100 m.
NO, This is not an example of work. The force is upward on the box but the displacement is along the ground.

6. You pull your luggage on a cart that makes an angle of 30º for 5 m
Yes, but only in direction of the displacement
5 m
So the force that does the work is the component of the force along the ground or Fx
So Work = Fd = (Fcos Θ)d

7. W = F d W = (Newtons )(meters) W = Nm W = Joule (J)
The units of work are;
W = F d
W = (Newtons )(meters)
W = Nm
W = Joule (J)
In customary; W = foot pounds

8. Matt lifts a 80 kg barbell upward for 1 meter at a constant speed, how much work does he do?
What force must Matt provide ?
F = w = mg = (80 kg) (10m/s/s) = 800 N
W= Fd = (800 N) (1 m) =
800 J

9. Vertical F = 200 N, d = 10 m W= Fd = (200 N) (10 m) = 2000 J
Ben carries a 200-N suitcase up three flights of stairs (a height of 10.0 m) and then pushes it with a horizontal force of 50.0 N at a constant speed of 0.5 m/s for a horizontal distance of 35.0 meters. How much work does Ben do on his suitcase during this entire motion?
Vertical
F = 200 N, d = 10 m
W= Fd = (200 N) (10 m) = 2000 J
F = 50.0 N, d = 35.0 m
Horizontal
W= Fd = (50.0 N) (35 m) = 1750 J
Total work = 3750 J

10. What work is done by a 60 N force in dragging the bag a distance of 50 m when the force is transmitted by a handle making an angle of 30 with the horizontal?
F = FcosΘ d = 50m
W= F∙d
W= FcosΘd
F = 60 N
W= (60 N)(cos 30º)(50m) Θ
FcosΘ
W= J

11. POWER is the rate at which work is done.
Power = (work)
(time)
P = W t
P = Joules
sec
P = J/s = Watts = W
In customary; Power = horsepower= hp
760 W = 1 hp

12. What is the man’s power in lifting a 3
What is the man’s power in lifting a 3.0 kg object through a vertical distance of 1.6 m in 10 sec?
F = w = mg
d = 1.6 m
t = 10 s
P = W = Fd
t t
P = (3 kg) (10 m/s/s) (1.6 m)
10 sec
P = 4.8 W

13. WORK, POWER, & ENERGY part 2

14. Energy is the ability to do work or that which can be converted into work..
Pg. 66
When something has energy, it is able to perform work or, in a general sense, to change some aspect of the physical world

15. In mechanics we are concerned with two kinds of energy:
KINETIC ENERGY: KE, energy possessed by a body by virtue of its motion.
KE = ½ mv2
Units: Joules (J)
POTENTIAL ENERGY: PE, energy possessed by a system by virtue of position or condition.
PE = m g h
Units: Joules (J)

16. Example: Find the kinetic energy of a 3200 N automobile traveling at 20.8 m/s?
Fg = 3200 N
v = 20.8 m/s
m = W/g = 320 kg
KE = ½ mv2
= ½ (320 kg) (20.8m/s)2
= 6.92 x104 J

17. Example: A 250 g object is held 200 mm above a workbench that is 1 m above the floor. Find the potential energy relative to
a. the bench top
m = 0.25 kg
h = 0.2 m
PE = mgh
= 0.25 kg (10 m/s2) (0.2m)
= 0.50 J
b. the floor
h = 1.2 m
PE = mgh
= 0.25 kg (10 m/s2) (1.2m)
= 3.00 J

18. W =  KE = KEf –KEo WORK-ENERGY PRINCIPLE:
The work of a resultant external force on a body is equal to the change in kinetic energy of the body.
W =  KE = KEf –KEo

19. W = ΔKE = ½ mvf2 - ½ mvo2 Fd = ½ mvf2 – ½ mvo2 F = - ½ mvo2 d
Example What average force F is necessary to stop a 16 g bullet traveling at 260 m/s as it penetrates into wood at a distance of 12 cm?
W = ΔKE = ½ mvf2 - ½ mvo2
m = kg
d = 0.12 m
vo = 260 m/s
vf = 0 m/s
Fd =
½ mvf2 – ½ mvo2
F = - ½ mvo2 d
F = N

20. W = PE W = mg(hf-ho) WORK-ENERGY PRINCIPLE Part II:
The work of a resultant external force on a body is equal to the change in gravitational
potential energy of the body.
W = PE
W = mg(hf-ho)

21. LAW OF CONSERVATION OF ENERGY
The law of conservation of energy states that:
"Energy is neither created nor destroyed."
Energy can be transformed from one kind to another,
but the total amount remains constant.

22. For mechanical systems involving conservative forces, the total mechanical energy equals the sum of the kinetic and potential energies of the objects that make up the system.

25. For mechanical systems involving conservative forces, the total mechanical energy equals the sum of the kinetic and potential energies of the objects that make up the system.

26. Einitial = Efinal KEi + PEi = KEf + PEf mgh = ½ mv2 = 5.66 m/s
Example A 40 kg pendulum ball is pulled to one side until it is 1.6 m above its lowest point. What will its velocity be as it passes through its lowest point?
m = 40 kg
h = 1.6 m
h=0
Einitial = Efinal
KEi + PEi = KEf + PEf
mgh = ½ mv2
= 5.66 m/s

27. At point A Energy: PEA + KEA At point B PEA + KEA = KEB +PEB
Example If friction forces are negligible and the coaster has a speed of 2 m/s at point A,
a). What will be its speed at point B? A
vA = 2 m/s
hA = 0.8 m
hB = 0 m C B
At point A
Energy: PEA + KEA
At point B
PEA + KEA = KEB +PEB
= 4.47 m/s

28. At point C PEA + KEA = PEC + KEC = 3.16 m/s
b. What will be its speed at point C?
hC = 0.5 m A C
At point C
PEA + KEA = PEC + KEC B
= 3.16 m/s

29. PSE Practice Exercises # 1-10
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