Chapter 5 Work and Energy
Chapter 5 Work and Energy
Chapter 5 Work and Energy
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
Θ= 0 Cos(0) = 1
Θ= 180 Cos(180) = -1
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
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
W= Fdcosθ= (24.62N)(253m)(cos180)
= -6230 J
Find μk
What is FN?
Ff
Fn + Fapp,y = Fg k
FN
Fn = Fg - Fapp,y= 70N- 40sin(52)= 38.48 N
Ff24.63
k .64
FN 38.48
Energy- Section 5.2 p. 172
Kinetic Energy- The energy of an object due
to its motion
1 2
KE mv
2
Kinetic energy = ½ x mass x speed2
2 KE 2(31.5 J ) m
v 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
1 2 1 2
Wnet Fnet d cos KE mv f mvi
2 2
1 2 1 2 1 1 2
mv f mvi 2000 2 2
( 2000)( 0 )
2 2 2 2
d 21m
F net cos 190 cos0
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
1 2 1
kx 51.0.035 .031J
2
PEelastic
2 2
Conservation of Energy – 5.3
The total amount of energy in the universe is
a constant
ME= KE + PE
Conservation of ME
In the absence of
friction, mechanical
energy is conserved
MEi ME f
When friction is present,
ME can be converted to
other forms of energy
(i.e. thermal energy) so
it is not conserved.
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
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 2 1 2
mvi mghi mv f mgh f
2 2
1
mv 2f mghi mgh f
2
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 hf = 0
1 Watt = 1 J
1s
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