Chapter 6: Work and Kinetic Energy
•  What is work done by a force
•  What is kinetic energy
•  work-energy theorem
•  How to calculate work done by a varying
force along a curved path
•  The meaning and calculation of power in a
physical situation
What does the work do (mean): The work-energy theorem
•  Work done on an object can change its motion and energy.
•  Relate these three topics we’ve studied so far:
•  Newton’s second law: F=ma
•  Definition of work:
W=Fs
•  Straight line motion (constant acceleration)
•  v22=v12+2as
1 2 1 2
W = Fs = mv 2 − mv1
2
2
1 2 In later chapter, this
• Definition of kinetic energy: K = mv
is related to energy
2
conservation!
W
=
K
−
K
=
ΔK
2
1
€• Work-energy theorem: tot
work done by the net force equals the change in kinetic energy
€
€
Work, a force through a distance
•  As in the illustration, the pushing (force) is in the same
direction that the object moves
•  The work is only “done” by a force when the force
contributes to the motion


•  What if F and s are not in the same direction?
• Unit: joule
1J =1N • m
€
Use the parallel component if the force acts at an angle
•  Only the component parallel to the displacement does the work
 
W = F • s = Fscos φ
•  φ: the angle between

F, s
Scalar product
•  If the force and displacement are known by their compoenents
F = Fx iˆ + Fy ˆj = (160N ) iˆ − (40N ) ˆj

s = xiˆ + yˆj = (14m)iˆ + (11m) ˆj
•  then
€
W = Fx x + Fy y
• Example 6.1 (b)
Scalar product of two vectors
• Numbers can be multiplied; What about vectors
• It can be multiplied: Scalar product and cross product
Dot product
• (Available in CH1 vectors in the text book)
 
 
A • B = ABcos φ =| A || B | cos φ
• What else do you need to know?
• What if φ=90o? φ=0o? φ=180o
• Using components (2d)
 
A • B = Ax Bx + Ay By
• It is a Scalar, and has no direction!
€
€
   
A• B = B• A
iˆ • iˆ = 1
iˆ • ˆj = 0
How to find angles between two vectors: from 2d to 3d
How to
find the angle between two vectors in 2d?



A = 2i + 2 j



B = −2 i + 2 j
3d too difficult? Use scalar products:
€
 
 
A • B = Ax Bx + Ay By + Az Bz =| A || B | cos φ
Ax Bx + Ay By + Az Bz
 
⇒ cos φ =
| A || B |
• Example 1.11


 
A = 2i + 3 j + k


 
B = −4 i + 2 j − k
How can it be such a great “workout” with no work?
•  When positive and negative work cancel, the net work
is zero even though muscles are exercising.
Q6.1
An elevator is being lifted at a constant
speed by a steel cable attached to an electric
Cable
motor. Which statement is correct?
A. The cable does positive work on the
elevator, and the elevator does positive
work on the cable.
Motor
v
Elevator
B. The cable does positive work on the elevator, and the elevator
does negative work on the cable.
C. The cable does negative work on the elevator, and the
elevator does positive work on the cable.
D. The cable does negative work on the elevator, and the
elevator does negative work on the cable.
Q6.2
An elevator is being lowered at a constant
speed by a steel cable attached to an electric
Cable
motor. Which statement is correct?
A. The cable does positive work on the
elevator, and the elevator does positive
work on the cable.
Motor
v
Elevator
B. The cable does positive work on the elevator, and the elevator
does negative work on the cable.
C. The cable does negative work on the elevator, and the
elevator does positive work on the cable.
D. The cable does negative work on the elevator, and the
elevator does negative work on the cable.
Work done by multiple forces
W tot
 
 
= (∑ F ) • s ≡ ∑ ( F • s )
• Work done by the net force
€
• Wtot=?
• Sum of the work done by each force
We can compare the kinetic energy of different bodies
•  Changes in the energy
of a moving body
under the influence of
an applied force
change differently
depending on the
direction of
application.
• What is K1:K2:K3?
K1
K2
K3
We can compare the kinetic energy of different bodies
•  Changes in the energy
of a moving body
under the influence of
an applied force
change differently
depending on the
direction of
application.
K1
K2
1
1
1
K1 : K 2 : K 3 = mv 2 : (2m)v 2 : m(2v) 2 = 1: 2 : 4
2
2
2
Ratio problem like this is your
professor’s favorite multiple choice question
K3
Q6.4
A tractor driving at a
constant speed pulls a
sled loaded with
firewood. There is
friction between the
sled and the road.
The total work done on the sled after it has moved a distance d is
A. positive.
B. negative.
C. zero.
D. not enough information given to decide Q6.5
A nonzero net force acts on an object. Which of the following
quantities could be constant?
A. the object’s kinetic energy
B. the object’s velocity
C. both of the above
D. none of the above
Using Work-Energy theorem, to find speed
•  Example 6.3.
From example 6.2:
• Wtotal = 10kJ
• Weight=14700N
• Can be done using acceleration too. Result should be the same!
• Please perform the exercise using the acceleration to find the speed
youself!
Application of work-energy theorem
• A baseball travelling at 20m/s.
•  Mike catches it with his hand, and makes it stop
• How does these two relate to each other:
• The distance (s) Mike draw his hand back in order to stop the ball
• The force (F) Mike receives from the ball?
A. The larger s is, the smaller F is
B. The larger s is, the larger F is
Different objects, different kinetic energies
•  Example:
–  Two boats, same sail
–  Different mass
–  Receive same force
•  Question:
–  Who cross the line first
–  Who has larger kinetic energy at finish line
–  What’s the ratio of their speed at finish line?
Work and energy with varying forces—Figure 6.16
•  Example: driving a
car, alternating
your attention
between the gas
and the brake.
•  The effect: positive
or negative force of
various magnitude
along a straight
line.
W =
∫
x2
x1
Fx dx
The stretch of a spring and the force that caused it
•  The force applied to an
ideal spring will be
proportional to its
stretch.
•  The graph of force on
the y axis versus stretch
on the x axis will yield a
slope of k, the spring
constant.
•  What’s the work done on
the spring?
Stepping on a scale (example 6.6)
•  Whether you like the result or not, stepping on a scale is an
excellent example of applied force and the work being done to
compress that spring.
•  Weight=600N; what is the spring constant k and total work done
on it, by you, during the compression?
Stepping on a scale
Fspring (Δx = −1.0cm) = k | Δx |= w
w
600N
k=
=
= 60000N /m
| Δx | 0.010m
1
1
1
W you _ on _ spring = kx 22 − kx12 = k((−0.01m) 2 − 0) = 3.0J
2
2
2
€
Motion with a varying force
•  Glider has intial speed
of v1=1.50m/s
•  m=0.100kg, k=20.0N/m
•  What is the maximum
distance dmax? Assuming
µ=0
•  What is the maximum
distance dmax? Assuming
µ=0.47
Motion with a varying force
•  Glider has intial speed
of v1=1.50m/s
•  m=0.100kg, k=20.0N/m
•  What is the maximum
distance dmax? Assuming
µ=0
•  What is the maximum
distance dmax? Assuming
µ=0.47
1 2
1 2
W = − kd = 0 − mv1
2
2
€
W tot = W spring +W f
1
1 2
2
W tot = − kdmax − µmgdmax = 0 − mv1
2
2
€
Motion on a curved path
•  Pushing a boy on a swing very slowly to angle θ0
•  Boy remain nearly in equlibrium.
•  What is the work done by total force?
•  What is the work done by the tension force
•  What is the work done by the gravity


W w = ∫ w • dl
dl = Rdθ
⇒ Ww =
∫ wRcos φdθ
φ = 90 o + θ
⇒ Ww =
∫
θ0
0
WF + Ww = 0
wR(−sin θ )dθ = wR(cosθ 0 −1)
Example:
• Block B moves down 75.0cm at constant speed
• How much work is done on B
• By gravity
• By the tension in the string
• By the total force?
• How much work is done on A
• By gravity
• By the tension in the string
• By friction
A
• By the normal force
•  By the total force
B
Q6.9
Two blocks are connected as
shown. The rope and pulley are of
negligible mass. When released,
the block of mass m1 slides down
the ramp and the block of mass m2
moves upward.
After each block has moved a distance d, the total work done on m1 A. is greater than the total work done on m2.
B. is the same as the total work done on m2.
C. is less than the total work done on m2.
D. not enough information given to decide Watt about power?
•  Power: time rate of doing work
Pav =
ΔW
Δt
•  Unit: Watt (James Watt)
•  Kilowatt-hour=?J
€
•  It’s work/energy, not Power!
•  Instantaneous power:
dW
P =
dt
•  Power and velocity:
F// Δs
Δs
= F// € = F//v av
Δt
Δt
Δt →0 :
 
p = F//v = F • v
Pav =
• 1 hp=?k W
Let’s power climb
•  Climbing the 443-m-tall Sears Tower in 15.0 minutes
•  m=50.0kg
pav =
•  What’s the average power
W = mgh (50.0)(9.8)(443)
=
= 241W
t
60 × 15
or
h
pav = F//v av = (mg) = 241W
t
output by her?
€
Summary
   
• Work (Joul, J) done by a force W = F • s ≡| F || s | cos φ
• Work done by multiple forces W
tot
1 2
• Kinetic energy: K = mv
2
 
 
= (∑ F ) • s ≡ ∑ ( F • s )
€ W tot = K 2 − K1 = ΔK
• Work-energy theorem:
€
• Work done by varying force:
€
• Work done on€a curved path:
ΔW
• Power (Watt, W): Pav =
Δt
€
dW
P =
dt
€
€
€
W =
W =
∫
∫
x2
x1
P2
P1
 
p=F•v
Fx dx
 
Fdl