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Honors Physics
WORK & ENERGY
Chapter 5
Giancoli 6th edition
ENERGY comes in a variety of forms
Energy: is the capacity to do work
Kinetic
Potential
 Chemical
 Elastic
 Heat
 Mechanical
 Nuclear
 Electric

Work
Energy is not lost, it is transformed
into other types of energy or work
KE
Work done by Constant Force
WORK
Force
PE
Energy
Specific meaning in physics
WORK
Work is the result of applying a force to a mass multiplied by the displacement of the mass as a
result of that force
WORK


When the force is stopped being applied, the work being done stops.
When the displacement stops the work being done also is discontinued
.
W = F║•d
W = cosF•d
F
displacement
W = the parallel force x displacement
W = F•d = N•m = J (joule)
Do not confuse force with work
Example 1 A force of 50 N pushes a wooden block 5.0 m. How much work was done to the wood
block?
W = 50N•5.0 m = 250J
Example 2 A 50 N force is pulling a 12 kg block at an angle of 30o as shown. If the block moves 10 m
how much work is done to the box?
30o
W = cos 30o (50N) 10m = 433J
Work done by a variable Force
Energy: is the capacity to do work
Elastic energy
PE = ½ kx2 k = spring constant
Hooke’s Law F=Kx
spring rating
K = F/x N/m
A spring has a 15 N weight placed on it and it stretches 12 cm.
K =15N/.12cm = 125N/m
Et = PE + KE = ½ kx2 + ½ mv2
Work- Kinetic Energy Principle ( kinetic energy)
NET WORK done to an object will change the object’s KE
KE = ½ mv2
vf2 = vo2
+ 2ad
a=
W = Fd = mad = m(
𝑣𝑓2 −𝑣𝑜2
𝑣𝑓2 −𝑣𝑜2
2𝑑
2𝑑
)d = m
1
𝑣2
2 𝑓
1
− 2 𝑣𝑜2 = ½ m(𝑣𝑓2 − 𝑣𝑜2 )
Work – Energy Theorem
Wnet = KE = KEf- KEo
Work is a measure of the transfer of KE to or from an object
 Work is something that is done to an object
 Energy is something that an object has, the ability to do work
Work & energy types both use the same units (joules) They are related because they both can be
expressed as Nm (joules)
Work is the link between PE and KE
To change a force into energy, work must first be done by the force.
PE
KE
work
work
kinetic energy
potential energy
An object that has KE has the ability to do work
A moving object may do work to a second object, It will exert a force on the
second object and cause a possible displacement which results in a change in
KE
Example: Hammer & Nail
Force from a hammer causes a nail to move ( KE) positive work = + KE
Force from the nail caused the hammer to stop (KE) negative work = -KE
Example: A 1200 kg car changes its velocity from 20 m/s to 30 m/s. How much work was done to the
car?
W= KE = ½ 1200 kg ( 30m/s)2 – ½ 1200kg (20m/s)2 = 540,000 J – 240,000 J = 300,000J
How much force would be needed to generate this change in velocity in 50 m?
W= F•d , F•50m = 300,000J
F = 300,000J/50m = 6,000N
Sometimes called the energy of position
Potential Energy (U)
is work)
Examples
Gravitational
Elastic
Chemical
Atomic
Thermal
Energy that is stored and available to do work or change into KE (which
GRAVITATIONAL PE (U)
A mass’s gravitational PE is due to its weight and position
PEgravity = mgh or mgy
The distance h (y) is measured against a reference point
 Reference point is considered to be zero = Uo point of zero PE
o Reference point is completely arbitrary
o Difference in PE between two points is the same regardless of the reference
point
 ΔPE = PEf - PEo
Example
A 25 kg mass is raised 20 m above a side walk. What is its gravitational PE?
PEg = mgh = 25kg•(9.8m/s2)•20m
sidewalk is the reference point
= 4900J
gravitational PE is part of a system. The mass and the body providing the gravity
Elastic potential energy
U = ½ kx2 k = spring constant
spring rating
Hooke’s Law F=Kx K = F/x N/m
A spring has a 15 N weight placed on it and it stretches 12 cm.
K =15N/.12cm = 125N/m
Et = PE + KE = ½ kx2 + ½ mv2
Sum of KE + PE in a system
CONSERVATION OF MECHANICAL ENERGY
Mechanical Energy; defined as the sum of the KE and PE (U) in a system

or Total Mechanical Energy = PE +  KE
o Et = KE1 + PE1 = KE2 + PE2
o Total Equation
 Et = KE1 + PE1 = KE2 + PE2 + LOST ENERGY

Conservation of physical quantity means that it has a constant value

In a closed (isolated) system the total quantity of energy remains constant
o Only truly isolated system is the universe
LAW of CONSERVATION OF ENERGY: The total amount of energy in a closed (isolated) system is
conserved.
 Energy may be converted from one form into another but the total amount of all forms of energy
remains constant. Total amount of energy remains unchanged.
Largest closed system imaginable is the universe.
Anything in this closed system can not interact with anything outside of the system
ENTROPY
CONSERVATIVE FORCES & NON-CONSERVATIVE FORCES
 Conservative Force depends on the initial and final positions of a mass.
o The concept of potential energy is only associated with potential energy
 Work done is equal to the potential energy gained
 An object lifted to a new height (mgy) has an increase in U and the work is
conserved as U
o Conservative forces can be recovered from the increase in U
o A force is conservative if the work it does when moving an object through a round trip is
zero.
 Non-conservative forces can not be recovered, it depends on the path taken to achieve stored U

Move an object from the floor to a table across the room
The gravitational PE can be recovered. The energy used to move the object across the
room is not recoverable so it is NOT CONSERVED
NON_CONSERVATIVE Forces often lose energy through friction or heat
PROBLEM SOLVING using CONSERVATION OF MECHANICAL ENERGY
Conservation of Gravitational Energy
Example: Carry a rock (m = 4.5 kg) to a height of 4.0 m above the ground and hold it there.
A. Determine the total mechanical energy of the rock
Et = KE1 + PE1
Et = ½ mv2 + mgh = ½ 4.5kg(0m/s)2 + 4.5kg(9.8m/s2)4.0m
= 0 + 176.4J = 176.4 J
B. Determine the velocity of the rock as it reaches the ground
Et = KE2 + PE2
176.4 J = ½ 4.5kg(v)2 + 4.5kg(9.8)(0m)
176.4 J = ½ 4.5kg v2 +0
v2 = 78.4m2/s2
v =8.85 m/s
C. Determine the velocity of the rock 1.0 m above the ground
Et = KE2 + PE2
176.4 J = ½ mv2 + mgh = ½ 4.5kg(v)2 + 4.5 kg (9.8)(1.0m)
176.4 J = ½ 4.5kg (v)2 + 44.1J
132.3 J = ½ 4.5kg v2
v2 = 58.8m2/s2
v = 7.77 m/s
A
Example 2
C
10m
5.0m
B
A 300kg roller coaster is traveling at 3m/s at pint A . Find the coaster’s velocity at
Point B
Point C
Et = KE1 + PE1 = ½ 300kg(3m/s)2 + 300kg(9.8)(10m) = 1350J + 29400J =30750J
Point B
Point C
Et = KE1 + PE1
30,750 J = ½ 300kg v2 + 300kg(9.8)0m
v2 = 205 m2/s2
v = 14.3 m/s
30,750 J = ½ 300kgv2 + 300kg(9.8) 5m
30,750 J = ½ 300kgv2 + 14,700J
16,050 J = ½ 300kg v2
v2 = 107m2/s2
v = 10.3 m/s
Conservation of Elastic energy
Et = PE + KE
PE = ½ kx2 k = spring constant
Hooke’s Law F=Kx
spring rating
K = F/x N/m
-X
0
+X
Example
A spring has a 15 N weight placed on it and it stretches 12 cm.
What is the velocity of the mass when it is at the + 9.0 cm position?
v12cm=?
k =?
ET =?
xmax = 12cm = 0.12m
m = wgt/9.8m/s2 = 15N=9.8m/s2 = 1.53kg
equation Et = PE + KE = ½ kx2 + ½ mv2
K =15N/.12cm = 125N/m
ET = ½ kx2 + ½ mv2 = ½ (125N/m)(0.12m)2 + ½(1.53kg)(0m/s)2
= 7.5J
ET = ½ kx2 + ½ mv2
7.5J = ½ (125N/m)(0.09m)2 + ½ (1.53kg)v2
v2 = 9.14m2/s2
v = 3.02 m/s = 3.0 m/s
POWER
Rate at which work is performed
P=
𝑤𝑜𝑟𝑘
𝑡𝑖𝑚𝑒
=
𝐹𝑑
𝑡
= F∙v =
Average P = 𝑃̅ =
𝐹(𝑐𝑜𝑠𝜃)𝑑
𝐹(𝑐𝑜𝑠𝜃)𝑑
P = 745.7 w = 1 hp
P = 0.746 kw = 1 hp
𝑡
𝑡
= F𝑣̅
P=
𝑤𝑜𝑟𝑘
𝑡𝑖𝑚𝑒
= Fvcosθ
=
𝐽
𝑠
= watts
MECHANICAL EFFICIENCY (ԑ)
ԑ=
𝑤𝑜𝑟𝑘 𝑜𝑢𝑡𝑝𝑢𝑡
𝑤𝑜𝑟𝑘 𝑖𝑛𝑝𝑢𝑡
ԑ = 𝑊𝑊𝑜𝑢𝑡
𝑖𝑛
ԑ=
=
40 𝐽
100𝑗
𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡
𝑃𝑜𝑤𝑒𝑟 𝑖𝑛
(𝑋 100) =
𝑊𝑜𝑢𝑡
𝑊𝑖𝑛
(𝑋 100)
x 100 = 0.40 (x 100) = 40 % efficiency
x 100 =
𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
x 100 =