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.
W  F d
W  K
P  W / T
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
Table 7-1
Typical Values of Work
Activity
Equivalent work (J)
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
Work is force times distance…but!
Only the force component in the direction of motion counts!
Units of work: Joule
Force in direction of motion is what
matters…
F
Fx = F cosq
D
W = D * F cosq
Figure 7-3
Force at an Angle to Direction of Motion: Another Look
I’ve got work to do….
F = Mg
F = Mg
W=F*y=MgH
y
This is the work done by the
person lifting the box.
How about the reverse?….
F = Mg
W=F*y
= - MgH
Work can be positive or negative.
Lifting a box is positive work.
Lowering the box is negative work.
H
F = Mg
Positive or Negative?
Work done by gravity.
Fg = - Mg
y = - H
Work “done by force of gravity” is
positive when an object is dropped.
Problem 7-77
W = 50J, given v, F, x.
What is theta?
What is M?
W  F (cos q )d
W
cos q 
Fd
W
q  cos 1
Fd
50.0 J
(45.0 N)(1.50 m)
 42.2
 cos 1
W  K 
m
2W
vf
2

1
1
m(vf 2  vi 2 )  mvf 2
2
2
2(50.0 J)
 2.60 
m
s
2
 14.8 kg
Figure 7-4
Positive, Negative, and Zero Work
Raindrops are falling on my
head….
W  F  y
 ( Mg )  ( H )
 MgH
V1
F = - Mg
y = -H
W  K

V2
MgH 
1
1
Mv22  Mv12
2
2
1
M (v22  v12 )
2
Special case: Object dropped from
rest.
Mass M dropped from height H. What
is speed just before hitting the
ground? (Neglect friction of air)
-Mg
W=MgH
H
W=K=1/2 M v2
v  2 gH
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.
How about the speed?
F=Mg sinq
H
q
If Block fell straight down through height H, its
speed would be:
v  2 gH
Use F=Ma to find speed down the ramp.
V=at
L=1/2 a t2
so
2L
va
a
 2aL
But a = g sinq
And L = H/sinq
So…
v  2 gH
Same magnitude of speed, but DIFFERENT DIRECTION!
Figure 7-6
Graphical Representation of the Work Done
by a Constant Force
Figure 7-7
Work Done by a Non-Constant Force
Figure 7-8a
Work Done by a Continuously Varying Force
Figure 7-8b
Work Done by a Continuously Varying Force
Figure 7-8c
Work Done by a Continuously Varying Force
When force is not constant.
The area under a force-distance
curve is the work, W.
Force
The average force times the
distance gives the work. This is
the same as the area under the
curve.
Distance
Back to our Spring Flings.
F = K x
F = -K x
Force
F=Kx
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.
W=½FX
W = ½ K X2
Distance
X
Work done to move a mass
on a spring a distance “x”:
W  12 Kx 2
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.
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.
Weightlifting Work.
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:
1. 1000J, 5000J, 1000J
2. 1000J, 0J, -1000J
3. 1000J, 5000J, -1000J
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.
What is the kinetic energy?
V = + 5 m/s
K = ½ M V2
M = 1 kg
V = -5 m/s
1.
2.
3.
4.
12.5 J
5J
Zero
Either zero or 12.5J,
depends on your frame of
reference.
The Power to overcome Friction.
Work is Force times distance. Power is Work per unit time.
AN IMPORTANT SPECIAL CASE
Block moves with STEADY SPEED under influence of
applied force that EXACTLY balances friction, so V is
CONSTANT.
Force of friction
Under these conditions,
F
P
W
F  X

 F v
T
T
Frictional Force, Ff = mN
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).
How about this case?
Total work is just the sum of the work for each spring.
1
1
K1 x 2  K 2 x 2
2
2
1
 K1  K 2 x 2
2
W
And this one?
This is harder. The two
springs DO NOT stretch the
same distance (they have
different K values).
x  x1  x2
F1  k1 x1
F2  k 2 x2
x 
But F1 = F2 (why?), so let it be called F.
F1 F2

k1 k 2
1 1
x  F   
 k1 k2 
So, can create an effective spring constant, Keff,
keff
Then, we can apply the usual formula: W 
1
k eff x 2
2


1


1  1 1

  
 1 1   k1 k 2 
k k 
2 
 1