Work and energy
1. Work
Definition:
W  Fd cos 
Work done by forces that oppose the direction of motion will be negative.
Units:
[W] = N*m = J
Example: A block slides down a rough inclined surface. The forces acting
on the block are depicted below. The work done by the frictional force is:
A. Positive
B. Negative
C. Zero
Wf = |fk| |Δx| cos(180°) = -|fk| |Δx| < 0
Work done by the normal force:
WN = |N| |Δx| cos(90°) = 0
N
f
Wy
y

Wx
x
W
Work done by weight:
Wmg = |mg| |Δx| cos(θ ) > 0
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2. Work kinetic energy principle
v22  v12  2a( x2  x1 )
m v22 m v12

 m a( x2  x1 )  Fd
2
2
Definition:
K
m v22 m v12
W

2
2
m v2
W=K2 - K1
2
Example: An 80-g arrow is fired from a bow whose string exerts
an average force of 100 N on the arrow over a distance of 49 cm.
What is the speed of the arrow as it leaves the bow?
m = 80 g
F = 100 N
d = 49 cm
v1= 0
v2 - ?
W  Fd
K1  0
m v22
K2 
2
m v22
Fd 
2
v2 
2 Fd
m
2 100N  49 102 m
v2 
 35m / s
3
80 10 kg
2
Example: Two blocks (m1=2m2) are pushed by identical forces, each
starting at rest at the same start line. Which object has the greater kinetic
energy when it reaches the same finish line?
1. Box1
2. Box 2
3. They both have the
same kinetic energy
Same force, same distance
Same work
Same change in kinetic energy
Example: A ball is dropped and hits the ground 50 m below. If the initial
speed is 0 and we ignore air resistance, what is the speed of the ball as it
hits the ground?
We can use kinematics or… the WKE theorem
Work done by gravity: mgh
W  K


mgh 12 mv2  0

v  2 gh  2 9.8m / s 2 50 m   31m / s
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3. Potential energy
U  Wconservative force
a) Gravitational potential energy:
U  m gh
F  mg
b) Elastic potential energy (spring):
kx 2
U 
2
F  kx
8. Conservation of energy
K1  U1  K 2  U 2
K2  K1  U 2  U1  0
K  U  0
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Example: A box of unknown mass and initial speed v0 = 10 m/s moves up a
frictionless incline. How high does the box go before it begins sliding down?
K1  U1  K 2  U 2
m
1
2
Only gravity does work (the normal
is perpendicular to the motion), so
mechanical energy is conserved.
mv02  0  0  mgh
2
v02

10m / s 
h

 5m
2
2 g 2  10m / s
We can apply the same thing to any “incline”!
Turn-around
point: where
K=0
E
K
U
E
K
U
E
K
U
v=0
h
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Example: A roller coaster starts out at the top of a hill of height h.
How fast is it going when it reaches the bottom?
Einitial  E final
U initial  mgh
K final
m v2
m gh 
2
m v2

2
h
v  2gh
Example: An object of unknown mass is projected with an initial speed,
v0 = 10 m/s at an unknown angle above the horizontal. If air resistance
could be neglected, what would be the speed of the object at height,
h = 3.3 m above the starting point?
v0  10m / s
h  3.3m
v ?
m v2
m v02
 m gh 
 m g0
2
2
v  v02  2 gh 
10m / s 2  29.8m / s 2 3.3m  6.0m / s
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Example: Pendulum (Conservation of energy)
Only weight of the pendulum is doing work; weight is a conservative force,
so mechanical energy is conserved:
K  U  const
θ0 θ0
L
m
K 0
U  max
K 0
U  max
U  m gh
m v2
K
2
K  max
U  min
The angle on
the other side
is also θ0!
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4. Energy in the simple harmonic motion
F   kx
U  12 kx2
E  12 mv2  12 kx 2
x  A cost   
v  A sin t   
U
E
–A
k  m 2
x
A
K
K  12 m 2 A 2 sin 2 t   
U  12 kA2 cos2 t   
E  kA  m A
1
2
2
1
2
2
t
U
t
2
E
t
Total mechanical energy is constant through
oscillation: conservation of energy!
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5. Damped Harmonic Motion
x(t)
Damping force is
proportional to velocity:
t
Fd  bv
b – damping constant
(Shows how fast
oscillations decay)
F  bv  kx  ma
Optional math:
x(t )  A(t ) cos( ' t   )
A(t )  A0 e t  A0 e b / 2 m t
 '   02  2 
k
  02
m
k  b 


m  2m 
2
b
 2
m
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6. Resonance
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