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Work and Energy
Physics Chapter 5
5.1 Work





Scientific definition of work: product of force
and displacement
Example: pushing a shopping cart
Equation: W = F d
Explain the only time work is done.
– Components of force is parallel to
displacement
Why is no work done by the student who is
carrying a bucket of water?
– Upward force exerted, moving horizontally
5.1 Work
Net Work Done by a Constant Net
Force
W = Fnet d (cos θ)
– Units: N (m) = Joules (J)
5.1 Work
Example 1: How much work is done
on a vacuum cleaner pulled 3.0 m by
a force of 50.0 N at an angle of 30.0o
above the horizontal?
W = Fd (cos θ)
W = 50.0 N (3.0 m) (cos 30o)
W = 130 J
5.1 Work
Example 2: A 20.0 kg suitcase is raised 3.0
m above a platform by a conveyor belt.
How much work is done on the suitcase?
Fg = mg
Fg = 20 kg (9.81 m/s2)
Fg = 196.2 N
W = Fd
W = 196 N (3.0 m)
W = 588 J
5.2 Kinetic Energy

Kinetic energy: energy associated
with an object in motion
Kinetic Energy
KE = ½ mv2
–
What type of quantity is kinetic energy?

–
Scalar (no direction = can’t be negative)
Units: Joule
5.2 Kinetic Energy


Kinetic energy depends upon what 2
things?
– Speed (has the greater impact on KE)
– Mass
Example: bowling ball and volleyball
– Traveling at same speed, bowling ball
has more KE because more mass
– Traveling at different speeds,
volleyball may have more KE if moving
a lot faster
5.2 Kinetic Energy
Example 3: A 6.0 kg cat runs after a
mouse at 10.0 m/s. What is the
cat’s kinetic energy?
KE = ½ mv2
KE = ½ (6.0 kg)(10.0 m/s)2
KE = 300 J
5.2 Kinetic Energy
Example 4: A 7.00 kg bowling ball moves
at 3.00 m/s. How much kinetic energy
does the bowling ball have? How fast
must a 2.45 g table-tennis ball move in
order to have the same kinetic energy as
the bowling ball? Is this speed reasonable
for a table-tennis ball?
Bowling ball
KE = ½ mv2
KE = ½ (7.0 kg)(3.0 m/s)2
KE = 31.5 J
5.2 Kinetic Energy
Table Tennis Ball
KE = ½ mv2
31.5 J = ½ (0.00245 kg)(v2)
31.5 J = 0.001225 (v2)
25714.29 = v2
v = 160 m/s
~345 mi/h
5.2 Work-Kinetic Energy
 Work-kinetic
energy theorem:
net work done by a net force acting
on an object is equal to the change
in KE
Work-Kinetic Energy Theorem
Wnet = ΔKE
Wnet = KEf – KEi
Wnet = ½mvf2 – ½mvi2
5.2 Work-Kinetic Energy
 Explain
how we can tell if the speed
of the object changes with the net
work done.
– Net work done is positive if speed
increases
 This
theorem allows us to think of
kinetic energy as:
– The work an object can do as it comes
to rest, or amount of energy stored in
the object
5.2 Work-Kinetic Energy
 Example:
– Moving hammer has KE and can do
work on the nail.
– The KE is transferred to work on the nail
– The hammer can only do as much work
as it has KE
 500
J of KE is converted to 500 J of work
5.2 Work-Kinetic Energy
Example 5: A 1200 kg car is accelerated from 15.0
m/s to 25.0 m/s in a distance of 50.0 m. What
net force acted upon the car?
Wnet = ½mvf2 – ½mvi2
W = ½(1200 kg)(25.0 m/s)2 – ½ (1200 kg)(15.0 m/s)2
W = 3.75 x 105 J – 1.35 x 105 J
W = 2.4 x 105 J
W = Fd
2.4 x 105 J = F(50.0 m)
F = 4.8 x 103 N
5.2 Work-Kinetic Energy
Example 6: On a frozen pond, a person
kicks a 10.0 kg sled, giving it an initial
speed of 2.2 m/s. How far does the sled
move if the coefficient of kinetic friction
between the sled and the ice is 0.10?
m = 10.0 kg
vf = 0 m/s
vi = 2.2 m/s
µk = 0.10
Fnet = µk (m)(g)
Fnet = 0.10 (10.0 kg)(9.81 m/s2)
Fnet = -9.81 N
5.2 Work-Kinetic Energy
W = ½mvf2 – ½mvi2
W = 0 J - ½ (10.0 kg)(2.2 m/s)2
W = -24.2 J
Wnet = Fnet (d)
-24.2 J = -9.81 N (d)
d = 2.47 m
5.2 Potential Energy
 Potential
energy: the amount of
energy stored in an object
 What
two things does the potential
energy depend upon?
– Properties of the object (ex: mass)
– Interaction with environment (ex:
gravity)
5.2 Potential Energy
 Gravitational
PE: PE associated
with an object due to the position of
object relative to Earth or some
gravitational source
Gravitational Potential Energy
PEg = mgh
5.2 Potential Energy
 Explain
how gravitational PE must be
measured.
– Relative to a zero level
– Example: volleyball
A
B
C
5.2 Potential Energy
Elastic PE: PE in a stretched/compressed
elastic object
 Relaxed length: length when no external
forces are acting on the spring

Elastic Potential Energy
PEelastic= ½ k x2
k = spring constant
x = how far stretched/compressed
5.2 Potential Energy
 Spring
constant: how resistant a
spring is to being compressed or
stretched
– Small constant: flexible spring
– Large constant: stiff spring
5.2 Potential Energy
Example 7: A 70.0 kg stuntman is attached to a bungee cord
with an unstretched length of 15.0 m. He jumps off a
bridge spanning a river from a height of 50.0 m. When he
finally stops, the cord has a stretched length of 44.0 m.
Assuming the spring constant of the bungee cord is 71.8
N/m, what is the total potential energy relative to the water
when the man stops falling?
PEtotal
PEtotal = PEel + PEg
PEtotal = ½ kx2 + mgh
= ½(71.8 N/m)(29.0 m)2 + 70.0 kg(9.81)(6.0 m)
PEtotal = 3.02 x 104 J + 4.1 x 103 J
PEtotal = 3.43 x 104 J
5.2 Potential Energy
Example 8: When a 2.0 kg mass is attached to a
vertical spring, the spring is stretched 10.0 cm
such that the mass is 50.0 cm above the table.
What is the total potential energy associated with
this mass relative to the table, if the spring
constant is 15.0 N/m?
PEtotal = PEel + PEg
PEtotal = ½ kx2 + mgh
PEtotal = ½(15.0 N/m)(0.10 m)2 + 2.0 (9.81)(0.50 m)
PEtotal = 0.075 J + 9.81 J
PEtotal = 9.89 J
5.3 Conservation of Energy
 Mechanical
energy: sum of kinetic
energy and all forms of potential
energy
 Describe all forms of energy in a
pendulum clock.
– KE when pendulum is swinging
– PEg when at different heights
– PEelastic with springs, gears, etc.
5.3 Conservation of Energy
 Equation:
ME = KE + PEg +PEelastic
 Nonmechanical energy: energy
like nuclear, chemical, internal, and
electrical
5.3 Conservation of Energy
Conservation of Mechanical Energy
MEi = MEf
KEi + PEg,i + PEel,i = KEf +PEg,f +PEel,f
½mvi2 + mghi + ½kxi2 = ½mvf2 + mghf + ½kxf2
5.3 Conservation of Energy
Example 9: Starting from rest, a child goes down a
frictionless slide from an initial height of 3.00 m.
What is her speed at the bottom of the slide?
Assume she has a mass of 25.0 kg.
MEi = MEf
KEi + PEgi = KEf + PEgf
PEgi = KEf
mghi = ½ mvf2
25.0 kg(9.81)(3.0 m) = ½ (25 kg)(vf2)
736 J = 12.5 (vf2)
vf = 7.67 m/s
5.3 Conservation of Energy
Example 10: A hurdler runs with a speed of 11.7 m/s. She
jumps over the hurdle with a speed of 1.18 m/s. How high
is she off the ground when going over the hurdle? (mass =
48.2 kg)
MEi = MEf
KEi + PEgi = KEf + PEgf
KEi = KEf +PEgf
½ mvi2 = ½ mvf2 + mghf
½ (48.2)(11.7)2 = ½ (48.2)(1.18)2 + 48.2(9.81)(hf)
3299 J = 33.56 J + 472.84 (hf)
3265.4 J = 472.84(hf)
hf = 6.9 m
5.3 Conservation of Energy
 When
is mechanical energy not
conserved? Total energy?
– Mechanical: if friction is present
– Total energy: always conserved
5.4 Power
 Power:
rate at which energy is
transferred; or the rate at which
work is done
Power
P=W
P = Fv
Δt
Units: Watts
5.4 Power

Explain what makes one machine different
from another.
– All machines will do the same work
– But in different time intervals

Example: Horsepower in engines
– To go from 0 to 60, more horsepower faster
time
– Lower HP, still get to 60 but in more time
5.4 Power
Example 11: A 193 kg curtain needs to be raised
7.5 m, at constant speed, in as close to 5.0 s as
possible. The power rating for three motors are
listed at 1.0 kW, 3.5 kW, and 5.5 kW. Which
motor is best for the job?
P=W
P = mgd
Δt
Δt
P = 193 kg(9.81)(7.5 m)
5.0 s
P = 2839 W
= 2.8 kW
5.4 Power
Example 12: Two horses pull a cart. Each exerts a
force of 250.0 N at a speed of 2.0 m/s for 10.0
minutes. Calculate the power delivered by the
horses. How much work is done by the two
horses?
P = Fv
P = 250 N (2.0 m/s)
P = 500 W x 2 horses = 1000 W
P = W/ Δt
1000 W = W/600 s
W = 6.0 x 105 J