Transcript Chapter 5 Work and Energy

Chapter 5
Work and Energy
Definition of Work
 There is a difference between the ordinary
definition of work and the scientific definition
of work

Ordinary Definition: To do something that
takes physical or mental effort

Scientific definition: Work is equal to the
magnitude of the applied force times the
displacement of an object
What is necessary for work to be
done?
 A force that causes displacement of an object
does work on the object

Therefore, work is not done on an object
unless the object is moved because of the
action of a force
Work is being done!!!
What is necessary for work to be
done?
 Work is done only when components of a
force are parallel to a displacement

When the force on an object and the object’s
displacement are perpendicular, no work is
done
No Work was Done!!!
Displacement
Force
What about forces at an angle?
 Only the component of force that is parallel to
the direction of the object’s displacement
does work.

Example: A person pushes a box across a
frictionless floor
FBD For the Box
 What part of the applied force is parallel to
the displacement?
FN
Displacement of Box
Fapp,x
Fapp,y
Fapp
Fg
General Equation For Work Done
W  Fd cos
Work= Component of Force that does the work x
displacement x cos (angle between the force vector and
the displacement)
Net Work done by a Constant Force
 If there are many constant forces acting on the
object, you can find the net work done by finding
the net force acting on the object
Wnet  Fnet d cos 
Net work= Net force x displacement x cos of the angle between them
Angles between vectors
Vector Orientation
Angle between
them
Θ= 90
Cos(θ)
Θ= 0
Cos(0) = 1
Θ= 180
Cos(180) = -1
Cos(90) =0
Units for Work
 The unit for Work is the Joule
 I J= 1 Nm
 One Joule = One Newton x One meter
The sign of work is important
 Work is a scalar quantity, but it can be
positive or negative
 Work is negative when the force is in the
direction opposite the displacement

For example, the work done by the frictional
force is always negative because the frictional
force is opposite the displacement
Sample Problem p. 193 # 10
 A flight attendant pulls her 70 N flight bag a
distance of 253 m along a level airport floor at
a constant velocity. The force she exerts is
40.0 N at an angle of 52.0° above the
horizontal. Find the following:



The work she does on the flight bag
The work done by the force of friction on the
flight bag
The coefficient of kinetic friction between the
flight bag and the floor
FBD
FN
Fapp
Fapp,y
Ff
Fapp,x
Fg
Work done by flight attendant
 Only the component of Fapp that is parallel to the
displacement does work.


Fapp,x is parallel to the displacement
Fapp,x = 40cos52 = 24.63 N
 Remember that in the W=Fdcosθ equation, θ
represents the angle between the force vector and
the displacement vector
Fapp,x
d
 W= (24.63 N)(253 m) cos(0)= 6230 J
Θ = 0°
Work done by friction
 The bag is moving at constant velocity, so
what is Ff?

Ff = Fapp,x= 40cos(52)= 24.62 N
Ff

d
Θ =180°
W= Fdcosθ= (24.62N)(253m)(cos180)
= -6230 J
Find μk
 What is FN?
 Fn + Fapp,y = Fg
k 
Ff
FN
 Fn = Fg - Fapp,y= 70N- 40sin(52)= 38.48 N
Ff
24.63
k 

 .64
FN 38.48
Energy- Section 5.2 p. 172
 Kinetic Energy- The energy of an object due
to its motion
I Have Kinetic
Energy
I don’t Have Kinetic
Energy
Kinetic Energy Depends on Speed and
Mass
1 2
KE  mv
2
 Kinetic energy = ½ x mass x speed2
 The unit for KE is Joules (J)
Sample Problem p. 173
 A 7.00 kg bowling ball moves at 3.0 m/s. How
much kinetic energy does the bowling ball
have? How fast must a 2.45 g tennis ball
move in order to have the same kinetic
energy as the bowling ball?
Solve the Problem
 KE= ½ mv^2= ½ (7kg)(3m/s)^2= 31.5 J
 How fast must 2.45 g ball move to have the
same KE?


Convert g to kg  2.45 g = .00245 kg
Solve for v
v
2 KE
2(31.5 J )
m

 160
m
.00245kg
s
Work-Kinetic Energy Theorem
 The net work done by a net force acting on an
object is equal to the change in kinetic energy of
the object
Wnet  KE  KE f  KEi
 You must include all the forces acting on the
object for this to work!
Sample problem p. 176 #2
 A 2000 kg car accelerates from rest under the
actions of two forces. One is a forward force
of 1140 N provided by the traction between
the wheels and the road. The other is a 950 N
resistive force due to various frictional forces.
Use the work-KE theorem to determine how
far the car must travel for its speed to reach
2.0 m/s.
What information do we have?
 M= 2000 kg
 Vi= 0m/s
 Vf= 2 m/s
Ffriction= 950 N
Fforward= 1140 N
What is the Work-Ke Theorem?
Wnet  KE  KE f  KEi
 Remember that:
Wnet  Fnet d cos 
1 2 1 2
KE  mv f  mvi
2
2
 So Expand the equation to this:
1 2 1 2
Fnet d cos   mv f  mvi
2
2
1 2 1 2
Fnet d cos   mv f  mvi
2
2
 Solve for Fnet

Fnet= Fforward- Ffriction= 190 N forward
 Rearrange the equation for d and plug in values
1
1 2 1 2 1
2
2



mv

mv
2000
2

(
2000
)(
0
)



f
i 
2
2
2
2

  21m
d
F netcos 
190 cos0
 Why is θ= 0 in the denominator? Because the net force is in the
same direction as the displacement.
Potential Energy  Section 5.2
 Potential Energy is stored energy
 There are two types of PE


Gravitational PE
Elastic PE
Gravitational Potential Energy
 Gravitational Potential Energy (PEg) is the
energy associated with an object’s position
relative to the Earth or some other gravitational
source

m
PEg  mgh  mass x 9.81 2 x height
s
Elastic Potential Energy
 Elastic Potential Energy (PEelastic) is the potential
energy in a stretched or compressed elastic object.
1 2
PEelastic  kx
2
 k= spring (force) constant
 X= displacement of spring
Displacement of Spring
Sample Problem p. 180 #2
 The staples inside a stapler are kept in place
by a spring with a relaxed length of 0.115 m.
If the spring constant is 51.0 N/m, how much
elastic potential energy is stored in the spring
when its length is 0.150 m?
What do we know?
 K = 51.0 N/m
 We need to get x, in order to use the equation for
elastic PE
 X is the distance the spring is stretched or
compressed
 Relaxed length is 0.115 m, stretched length is 0.150.
How much was it stretched?

0.150- 0.115 m= 0.035 m= x
Solve the problem
1 2 1
2
PEelastic  kx  51.0.035  .031J
2
2
Conservation of Energy – 5.3
 The total amount of energy in the universe is
a constant

So we say that energy is conserved
 From IPC: The Law of Conservation of
Energy: Energy can neither be created nor
destroyed
Mechanical Energy
 There are many types of energy (KE, PE,
Thermal, etc)
 We are concerned with Mechanical Energy
 Mechanical Energy is the sum of kinetic
energy and all forms of potential energy
 ME= KE + PE
Conservation of ME
 In the absence of
friction, mechanical
energy is conserved
 When friction is present,
ME can be converted to
other forms of energy
(i.e. thermal energy) so
it is not conserved.
MEi  ME f
Expanded Form of Conservation of
ME
 Without elastic PE
1 2
1 2
mvi  mghi  mv f  mgh f
2
2
 With elastic PE
1 2
1 2 1 2
1 2
mvi  mghi  kxi  mv f  mgh f  kx f
2
2
2
2
Practice Problem p. 185 #2
 A 755 N diver drops from a board 10.0 m
above the water’s surface. Find the diver’s
speed 5.00 m above the water’s surface. Find
the diver’s speed just before striking the
water.
What do we know?
 W= 755 N
 Initial height = 10 m
 Vi= 0 m/s
 There is no elastic PE involved.
Solve part a.
 What is the diver’s speed 5.0 m above the
water’s surface?
 M= Weight/g=76.96 kg
 Vi= 0m/s
 Initial height = 10 m
 Final Height = 5 m
Rearrange equation and solve for vf
Vi = 0 m/s
1
1
2
2
mvi  mghi  mv f  mgh f
2
2
1
mv 2f  mghi  mgh f
2
vf 
2mghi  mgh f 
m
2((77)(9.81)(10)  (77)(9.81)(5)
m

 10
77
s
Second Part
 What is the diver’s speed just before striking
the water?
 M= Weight/g=76.96 kg
 Vi= 0m/s
 Initial height = 10 m
 Final Height = 0 m
Finish the Problem
Vi = 0 m/s
1 2
1 2
mvi  mghi  mv f  mgh f
2
2
1 2
mghi  mv f
2
hf = 0
2mghi
m
vf 
 2 ghi  2(9.81)(10)  14
m
s
Power- Section 5.4
 Power:The rate at
which work is done
 The unit for power is
Watts
 1 Watt = 1 J
1s
W Work
P

t
Time
Alternate Form for Power
P  Fv  Force x Speed
Sample Problem (Not in book)
 At what rate is a 60 kg boy using energy when he
runs up a flight of stairs 10 m high in 8.0 s?
W Work
P

t
Time
 Time = 8 s
 What is work done?
 W=Fdcos(θ)
Solve the Problem
 What force does the boy apply to get himself
up the stairs?


F= Weight= mg= 588.6 N
d= 10m
 W= Fdcos(θ)=588.6N(10m)(cos(0))

W=5886 J
 P=W/t = 5886 J/ 8s= 735.8 Watts