Work, energy, and power are defined. Work is force times distance. Kinetic energy is equal to one-half mass times velocity squared. Power is the rate of doing work, defined as work per unit time. Several examples of calculating work are provided for different scenarios like lifting an object, springs, and overcoming friction. The key concepts of work, kinetic energy, and power are reviewed.

1. Work, Energy, Power
Chapter 7 in a nutshell
Work is Force times Distance.
The change in Kinetic Energy is equal to the work.
Power is Work per unit time.
T
W
P
K
W
d
F
W






/

2. New Concept:
Kinetic Energy
K = ½ m V2
Work causes a change in kinetic energy, so the units
are the same.
Units of Kinetic energy: Joule
But, from the definition of kinetic energy, the units are also,
1 Joule = 1 Kg-m2/s2

3. Table 7-1
Typical Values of Work
Annual U.S. energy use 8 x 1019
Mt. St. Helens eruption 1018
Burning one gallon of gas 108
Human food intake/day
Melting an ice cube
107
104
Lighting a 100-W bulb for 1 minute 6000
Heartbeat 0.5
Turning page of a book 10–3
Hop of a flea 10–7
Breaking a bond in DNA 10–20
Activity Equivalent work (J)

4. Work is force times distance…but!
Only the force component in the direction of motion counts!
Units of work: Joule

5. Force in direction of motion is what
matters…
D
F
Fx = F cosq
W = D * F cosq

6. Figure 7-3
Force at an Angle to Direction of Motion: Another Look

7. I’ve got work to do….
F = Mg
F = Mg
y
W=F*y=MgH
This is the work done by the
person lifting the box.

8. How about the reverse?….
F = Mg
F = Mg
H
W=F*y
= - MgH
Work can be positive or negative.
Lifting a box is positive work.
Lowering the box is negative work.

9. Positive or Negative?
Work done by gravity.
Fg = - Mg
y = - H
Work “done by force of gravity” is
positive when an object is dropped.

10. Problem 7-77
W = 50J, given v, F, x.
What is theta?
What is M?
1
1
(cos )
cos
cos
50.0 J
cos
(45.0 N)(1.50 m)
42.2
W F d
W
Fd
W
Fd
q
q
q 





 
2 2 2
f i f
1 1
( )
2 2
W K m v v mv
    
 
2 2
m
f
s
2 2(50.0 J)
14.8 kg
2.60
W
m
v
  

11. Figure 7-4
Positive, Negative, and Zero Work

12. Raindrops are falling on my
head….
V1
F = - Mg
V2
y = -H
MgH
H
Mg
y
F
W








)
(
)
(
2
1
2
2
2
1
2
1
Mv
Mv
K
W




)
(
2
1 2
1
2
2 v
v
M
MgH 


13. Special case: Object dropped from
rest.
-Mg
Mass M dropped from height H. What
is speed just before hitting the
ground? (Neglect friction of air)
H
W=MgH
W=K=1/2 M v2
gH
v 2


14. Thought provoking: The inclined plane.
Distance = h
Force = mg
W = mgh
Distance = L = h/sin q
Force = mg sin q
W = mgh
The work is the same in both cases. This is an example of
conservation of energy, which we will see much more of in the
future.

15. How about the speed?
H
If Block fell straight down through height H, its
speed would be:
gH
v 2

q
Use F=Ma to find speed down the ramp.
F=Mg sinq
V=at
L=1/2 a t2 so
aL
a
L
a
v
2
2


But a = g sinq
And L = H/sinq
So…
gH
v 2

Same magnitude of speed, but DIFFERENT DIRECTION!

16. Figure 7-6
Graphical Representation of the Work Done
by a Constant Force

17. Figure 7-7
Work Done by a Non-Constant Force

18. Figure 7-8a
Work Done by a Continuously Varying Force

19. Figure 7-8b
Work Done by a Continuously Varying Force

20. Figure 7-8c
Work Done by a Continuously Varying Force

21. When force is not constant.
The area under a force-distance
curve is the work, W.
Distance
Force
The average force times the
distance gives the work. This is
the same as the area under the
curve.

22. Back to our Spring Flings.
F = -K x
How much work is done to move the
block?
Be careful with the signs of the
forces!
The force to move the block must be
equal and opposite to the force of
the spring on the block.
F = K x
Force
Distance X
F=Kx
W = ½ F X
W = ½ K X2
Work done to move a mass
on a spring a distance “x”:
2
2
1
Kx
W 

23. Conceptual Checkpoints
• What is work, in physics terms?
• What is kinetic energy?
• What is power?
Work is force in the direction of motion.
Kinetic energy is ½ Mv2
Power is Work (or Energy) per Unit time.

24. Work and the Weightlifter
A weightlifter performs three steps.
What is the work done for each of the
steps? Assume the weight is 100 Kg,
the height of the lift is 1 M, and the
acceleration of gravity is 10 M/s^2.
Work done against gravity
to lift an object a height H is
W = MgH
W1: The weight is lifted up to the height of 1 M.
W2: The weight is held at 1M for 5 seconds.
W3: The weight is lowered down to the ground.
Cheryl Hawarth, USA gold medal.

25. Weightlifting Work.
1. 1000J, 5000J, 1000J
2. 1000J, 0J, -1000J
3. 1000J, 5000J, -1000J
W1: The weight is lifted up to the height of 1 M.
W2: The weight is held at 1M for 5 seconds.
W3: The weight is lowered down to the ground.
The work done W1, W2, W3 is:

27. The energy is kinetic, relatively speaking….
V = + 5 m/s
V = -5 m/s
Train moves to the right
with speed V. A baseball
is thrown to the left with
the same speed.
What is the kinetic energy
of the baseball? Its mass
is 1 kg.

28. What is the kinetic energy?
1. 12.5 J
2. 5 J
3. Zero
4. Either zero or 12.5J,
depends on your frame of
reference.
V = + 5 m/s
V = -5 m/s
K = ½ M V2
M = 1 kg

29. The Power to overcome Friction.
Work is Force times distance. Power is Work per unit time.
F
Frictional Force, Ff = mN
Block moves with STEADY SPEED under influence of
applied force that EXACTLY balances friction, so V is
CONSTANT.
AN IMPORTANT SPECIAL CASE
Force of friction
Under these conditions,
v
F
T
X
F
T
W
P 







THIS IS A SPECIAL CASE IN WHICH AN APPLIED FORCE RESULTS IN CONSTANT SPEED.
IT OCCURS BECAUSE THE NET FORCE IS ZERO (SUM OF ALL APPLIED FORCES).

30. How about this case?
Total work is just the sum of the work for each spring.
  2
2
1
2
2
2
1
2
1
2
1
2
1
x
K
K
x
K
x
K
W





31. And this one?
This is harder. The two
springs DO NOT stretch the
same distance (they have
different K values).
2
2
1
1
2
2
2
1
1
1
2
1
k
F
k
F
x
x
k
F
x
k
F
x
x
x








But F1 = F2 (why?), so let it be called F.











2
1
1
1
k
k
F
x
So, can create an effective spring constant, Keff, 1
2
1
2
1
1
1
1
1
1

























k
k
k
k
keff
Then, we can apply the usual formula: 2
2
1
x
k
W eff 
