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PHYSICS 220
Lecture 9
Work and Kinetic Energy
Lecture 9
Purdue University, Physics 220
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For this example:
vf2 = v02 + 2ax = 2Fx/m
mv2/2 = Fx = W (work)
[N]*[m] = J (joule)
Lecture 9
Purdue University, Physics 220
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Work and Energy
•Work: Transfer of Energy by Force
W = |F| |r| cos
W depends on the direction of the force relative to the
displacement
Lecture 9
Purdue University, Physics 220
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Work by Constant Force
• Only component of force parallel to direction of
motion does work!
F
 W = F r cos 
F
WF > 0: 0<  < 90 : cos() > 0
r
F
WF = 0:  =90 : cos() =0
r
F
WF < 0: 90<  < 270 : cos() < 0
r
r
F
Lecture 9

WF > 0: 0<  < 90 : cos() > 0
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Question
You are towing a car up a hill with constant velocity.
The work done on the car by the normal force is:
A) positive
B) negative
C) zero
N
V
T
Normal force is perpendicular to direction of
displacement, so work is zero.
Lecture 9
Purdue University, Physics 220
W
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Question
You are towing a car up a hill with constant velocity.
The work done on the car by the gravitational force is:
A) positive
B) negative
C) zero
N
V
T
Gravity is pushing against the direction of
motion so it is negative.
Lecture 9
Purdue University, Physics 220
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Question
You are towing a car up a hill with constant velocity.
The work done on the car by the tension force is:
A) positive
B) negative
C) zero
N
V
T
The force of tension is in the same direction as
the motion of the car, making the work positive.
Lecture 9
Purdue University, Physics 220
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iClicker
You toss a ball in the air. The work done by
gravity as the ball goes up is:
A) Positive
Lecture 9
B) Negative
C) Zero
Purdue University, Physics 220
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Work by Constant Force
• Example: You pull a 30 N chest 5 meters across the floor
at a constant speed by applying a force of 50 N at an angle of
30 degrees. How much work is done by the 50 N force?
W = T x cos 
= (50 N) (5 m) cos (30)
N
T
f
mg
= 217 Joules
50 N
30
Lecture 9
Purdue University, Physics 220
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Where did the Energy go?
• Example: You pull a 30 N chest 5 meters across the floor at a
constant speed, by applying a force of 50 N at an angle of 30 degrees.
• How much work did gravity do?
W = mg r cos 
= 30  5 cos(90)
=0
• How much work did friction do?
r
90
mg
x-direction: SF = ma
T cos(30) – f = 0
f = T cos(30)
W = f r cos(180)
= 50 cos(30)  5 cos(180)
= -217 Joules
Lecture 9
Purdue University, Physics 220
N
T
f
mg
f
r
180
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Work by Variable Force
Lecture 9
Purdue University, Physics 220
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Work by Variable Force
Spring: F
spring=
-k x
Work is the area under the F vs x plot
W by spring force = -1/2 k x2
Lecture 10
Purdue University, Physics 220
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Kinetic Energy: Motion
• Apply constant force along x-direction to a
point particle m
W = Fx x
= m ax x
= ½ m (vf2 – v02)
(
1 2
2
ax x = vx - vx0
2
)
• Work changes ½ m v2
• Define Kinetic Energy (energy of motion)
KE = ½ m v2
W =  KE
Lecture 9
Work-Kinetic Energy Theorem
Purdue University, Physics 220
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Falling Ball Example
• Ball falls a distance of 5 meters. What is
its final speed?
Only force/work done be gravity
Wg = m ½ (vf2 – vi2)
Fg h = ½m vf2
mgh = ½m
vf2
mg
Vf = sqrt( 2 g h ) = 10 m/s
Lecture 9
Purdue University, Physics 220
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Example: Block with Friction
• A block is sliding on a surface with an initial speed of 5 m/s. If the
coefficient of kinetic friction between the block and table is 0.4, how
far does the block travel before stopping?
N
y-direction: F=ma
N-mg = 0
N = mg
Work
WN = 0
Wmg = 0
Wf = f x cos(180)
= -mmg x
5 m/s
Lecture 9
f
y
x
mg
W =  KE
-mmg x = ½ m (vf2 – v02)
-mg x = ½ (0 – v02)
mg x = ½ v02
x = ½ v02 / mg
= 3.1 meters
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Work: Energy Transfer due to Force
• Force to lift trunk at constant speed
– Case a Ta – mg = 0 Ta = mg
– Case b 2Tb - mg =0 or Tb = ½ mg
• But in case b, trunk only moves ½
distance you pull rope
• Work is same in both!
Ta
Lecture 9
mg
Tb Tb
W = mgh
mg
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iClicker
A box is pulled up a rough (m > 0) incline by a ropepulley-weight arrangement as shown below. How
many forces are doing (non-zero) work on the box?
A) 0
Lecture 9
B) 1
C) 2
Purdue University, Physics 220
D) 3
E) 4
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Solution
Draw FBD of box:
T

Consider direction of
motion of the box

Any force not perpendicular
to the motion will do work:
v
N
f
N does no work (perp. to v)
T does positive work
f does negative work
3 forces
do work
mg
mg does negative work
Lecture 9
Purdue University, Physics 220
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