Physics 101

Lecture 8
Potential Energy and
Conservation
Assoc. Prof. Dr. Ali ÖVGÜN
EMU Physics Department
November 25, 2024
 Kinetic Energy and Work
❑ Kinetic energy associated with the motion of
an object 1 2
KE = mv
2
❑ Scalar quantity with the same unit as work
❑ Work is related to kinetic energy
1 2 1
mv − mv0 2 = ( Fnet cos  )x Units: N-m or J
2 2
xf
=  F  dr
xi

Wnet = KEf − KEi = KE

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Work done by a Gravitational Force
❑ Gravitational Force
◼ Magnitude: mg
1 2 1
Direction: downwards to the Wnet = mv − mv0
2

◼ 2 2

Earth’s center
❑ Work done by Gravitational
Force
W = F r cos  = F   r

Wg = mgr cos 
❑ Using the general symbol W for work, we
write this as 11/25/2024
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 Potential Energy
❑ Potential energy is associated with the
position of the object
❑ Gravitational Potential Energy is the
energy associated with the relative
position of an object in space near the
Earth’s surface
❑ The gravitational potential energy
PE  mgy
◼ m is the mass of an object
◼ g is the acceleration of gravity
◼ y is the vertical position of the mass
relative the surface of the Earth
◼ SI unit: joule (J)

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Extended Work-Energy Theorem
❑ The work-energy theorem can be extended to include
potential energy:
Wnet = KEf − KEi = KE
Wgrav ity = PEi − PEf
❑ If we only have gravitational force, then Wnet = Wgravity
KE f − KEi = PE i − PE f
KE f + PE f = PE i + KEi
❑ The sum of the kinetic energy and the gravitational
potential energy remains constant at all time and hence
is a conserved quantity
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 Extended Work-Energy
Theorem
❑ We denote the total mechanical energy by

E = KE + PE
❑ Since KE f + PE f = PE i + KEi

❑ The total mechanical energy is conserved and remains

the same at all times
1 2 1 2
mvi + mgyi = mv f + mgy f
2 2

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 Spring Force
❑ Involves the spring constant, k
❑ Hooke’s Law gives the force
 
F = −kd
◼ F is in the opposite direction of
displacement d, always back
towards the equilibrium point.
◼ k depends on how the spring
was formed, the material it is
made from, thickness of the
wire, etc. Unit: N/m.
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Potential Energy in a Spring
❑ Elastic Potential Energy: 1 2
PEs = kx
◼ SI unit: Joule (J) 2
◼ related to the work required to
compress a spring from its
equilibrium position to some final,
arbitrary, position x
❑ Work done by the spring
xf 1 2 1 2
Ws =  (−kx)dx = kxi − kx f
xi 2 2
Ws = PEsi − PEsf
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 Extended Work-Energy
Theorem
❑ The work-energy theorem can be extended to include
potential energy:
Wnet = KEf − KEi = KE
Wgrav ity = PEi − PEf Ws = PEsi − PEsf
❑ If we include gravitational force and spring force, then
Wnet = Wgravity + Ws
( KE f − KEi ) + ( PE f − PE i ) + ( PE sf − PE si ) = 0

KE f + PE f + PE sf = PE i + KEi + KEsi

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 Extended Work-Energy
Theorem
❑ We denote the total mechanical energy by
E = KE + PE + PE s

❑ Since ( KE + PE + PE s ) f = ( KE + PE + PE s ) i

❑ The total mechanical energy is conserved and remains

the same at all times
1 2 1 2 1 2 1 2
mvi + mgyi + kxi = mv f + mgy f + kx f
2 2 2 2

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Ex:1 A block projected up a
incline
❑ A 0.5-kg block rests on a horizontal, frictionless surface.
The block is pressed back against a spring having a
constant of k = 625 N/m, compressing the spring by
10.0 cm to point A. Then the block is released.
❑ (a) Find the maximum distance d the block travels up
the frictionless incline if θ = 30°.
❑ (b) How fast is the block going when halfway to its
maximum height?

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 a) A block projected up a
incline
❑ Point A (initial state): vi = 0, yi = 0, xi = −10cm = −0.1m
❑ Point B (final state): v f = 0, y f = h = d sin  , x f = 0

1 2 1 2 1 2 1 2
mvi + mgyi + kxi = mv f + mgy f + kx f
2 2 2 2
1 2
2
kxi = mgy f = mgd sin 
1
2 kxi
2
d=
mg sin 
0.5(625 N / m)( −0.1m) 2
=
(0.5kg )(9.8m / s 2 ) sin 30
= 1.28m
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b) A block projected up a
incline
❑ Point A (initial state): vi = 0, yi = 0, xi = −10cm = −0.1m
❑ Point B (final state): v f = ?, y f = h / 2 = d sin  / 2, x f = 0
1 2 1 2 1 2 1 2
mvi + mgyi + kxi = mv f + mgy f + kx f
2 2 2 2
1 2 1 2 h k 2
kxi = mv f + mg ( ) xi = v 2f + gh
2 2 2 m
h = d sin  = (1.28m) sin 30 = 0.64m
k 2
vf = xi − gh
m
= ...... = 2.5m / s
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 Types of Forces
❑ Conservative forces
◼ Work and energy associated
with the force can be recovered
◼ Examples: Gravity, Spring Force,
EM forces
❑ Non conservative forces
◼ The forces are generally
dissipative and work done
against it cannot easily be
recovered
◼ Examples: Kinetic friction, air
drag forces, normal forces,
tension forces, applied forces …

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 Conservative Forces
❑ A force is conservative if the work it does on an
object moving between two points is
independent of the path the objects take
between the points
◼ The work depends only upon the initial and final
positions of the object
◼ Any conservative force can have a potential energy
function associated with it
◼ Work done by gravity Wg = PEi − PE f = mgyi − mgy f
◼ Work done by spring force 1 2 1 2
Ws = PEsi − PEsf = kxi − kx f
2 2
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 Non conservative Forces
❑ A force is non conservative if the work it does
on an object depends on the path taken by the
object between its final and starting points.
◼ The work depends upon the movement path
◼ For a non-conservative force, potential energy can
NOT be defined
◼ Work done by a non conservative force
 
Wnc =  F  d = − f k d + Wotherforces

◼ It is generally dissipative. The dispersal

of energy takes the form of heat or sound
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 Extended Work-Energy
Theorem
❑ The work-energy theorem can be written as:
Wnet = KEf − KEi = KE
Wnet = Wnc + Wc
◼ Wnc represents the work done by non conservative forces
◼ Wc represents the work done by conservative forces
❑ Any work done by conservative forces can be accounted
for by changes in potential energy W = PE − PE
c i f

◼ Gravity work Wg = PE i − PE f = mgyi − mgy f

1 2 1 2
◼ Spring force work Ws = PEi − PE f = kxi − kx f
2 2
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Extended Work-Energy Theorem
❑ Any work done by conservative forces can be accounted
for by changes in potential energy
Wc = PEi − PE f = −( PE f − PEi ) = −PE
Wnc = KE + PE = ( KE f − KEi ) + ( PE f − PEi )
Wnc = ( KE f + PE f ) − ( KEi + PEi )
❑ Mechanical energy includes kinetic and potential energy
1 2 1 2
E = KE + PE = KE + PEg + PEs = mv + mgy + kx
2 2
Wnc = E f − Ei

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 Problem-Solving Strategy
❑ Define the system to see if it includes non-conservative
forces (especially friction, drag force …)
❑ Without non-conservative forces
1 2 1 1 1
mv f + mgy f + kx 2f = mvi2 + mgyi + kxi2
2 2 2 2
❑ With non-conservative forces W = ( KE + PE ) − ( KE + PE )
nc f f i i

1 1 1 1
− fd + Wotherforces = ( mv 2f + mgy f + kx 2f ) − ( mvi2 + mgyi + kxi2 )
2 2 2 2
❑ Select the location of zero potential energy
◼ Do not change this location while solving the problem
❑ Identify two points the object of interest moves between
◼ One point should be where information is given
◼ The other point should be where you want to find out something

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 Ex2: Conservation of Mechanical
Energy
❑ A block of mass m = 0.40 kg slides across a horizontal
frictionless counter with a speed of v = 0.50 m/s. It runs into
and compresses a spring of spring constant k = 750 N/m.
When the block is momentarily stopped by the spring, by
what distance d is the spring compressed?
Wnc = ( KE f + PE f ) − ( KEi + PEi )

1 2 1 1 1 1 2 1 2
mv f + mgy f + kx 2f = mvi2 + mgyi + kxi2 0 + 0 + kd = mv + 0 + 0
2 2 2 2 2 2
1 1
0 + 0 + kd 2 = mv 2 + 0 + 0
2 2
m 2
d= v = 1.15cm
k

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 Ex 3: Block-Spring Collision
❑ A block having a mass of 0.8 kg is given an initial velocity vA = 1.2
m/s to the right and collides with a spring whose mass is negligible
and whose force constant is k = 50 N/m as shown in figure. Assuming
the surface to be frictionless, calculate the maximum compression of
the spring after the collision.
1 2 1 1 1
mv f + mgy f + kx 2f = mvi2 + mgyi + kxi2
2 2 2 2

1 2 1 2
mvmax + 0 + 0 = mv A + 0 + 0
2 2

m 0.8kg
xmax = vA = (1.2m / s ) = 0.15m
k 50 N / m

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 Block-Spring Collision
❑ A block having a mass of 0.8 kg is given an initial velocity vA = 1.2
m/s to the right and collides with a spring whose mass is negligible
and whose force constant is k = 50 N/m as shown in figure. Suppose
a constant force of kinetic friction acts between the block and the
surface, with µk = 0.5, what is the maximum compression xc in the
spring.
1 1 1 1
− fd + Wotherforces = ( mv 2f + mgy f + kx 2f ) − ( mvi2 + mgyi + kxi2 )
2 2 2 2
1 1
−  k Nd + 0 = (0 + 0 + kxc2 ) − ( mv A2 + 0 + 0)
2 2
N = mg and d = xc
1 2 1 2
kxc − mv A = −  k mgxc
2 2
25 xc2 + 3.9 xc − 0.58 = 0 xc = 0.093m
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Changes in Mechanical Energy for conservative
forces
❑ A 3-kg crate slides down a ramp. The ramp is 1 m in length and
inclined at an angle of 30° as shown. The crate starts from rest at the
top. The surface friction can be negligible. Use energy methods to
determine the speed of the crate at the bottom of the ramp.
1 1 1 1
− fd + Wotherforces = ( mv 2f + mgy f + kx 2f ) − ( mvi2 + mgyi + kxi2 )
2 2 2 2
1 1 1 1
( mv 2f + mgy f + kx 2f ) = ( mvi2 + mgyi + kxi2 )
2 2 2 2

d = 1m, yi = d sin 30 = 0.5m, vi = 0

y f = 0, v f = ?
1
( mv 2f + 0 + 0) = (0 + mgyi + 0)
2
v f = 2 gyi = 3.1m / s

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Changes in Mechanical Energy for Non-
conservative forces
❑ A 3-kg crate slides down a ramp. The ramp is 1 m in length and
inclined at an angle of 30° as shown. The crate starts from rest at the
top. The surface in contact have a coefficient of kinetic friction of 0.15.
Use energy methods to determine the speed of the crate at the bottom
of the ramp.
1 1 1 1
− fd + Wotherforces = ( mv 2f + mgy f + kx 2f ) − ( mvi2 + mgyi + kxi2 )
2 2 2 2
1 2 N
−  k Nd + 0 = ( mv f + 0 + 0) − (0 + mgyi + 0)
2
fk
 k = 0.15, d = 1m, yi = d sin 30 = 0.5m, N = ?
N − mg cos  = 0
1 2
−  k dmg cos  = mv f − mgyi
2
v f = 2 g ( yi −  k d cos  ) = 2.7m / s
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Changes in Mechanical Energy for Non-
conservative forces
❑ A 3-kg crate slides down a ramp. The ramp is 1 m in length and
inclined at an angle of 30° as shown. The crate starts from rest at the
top. The surface in contact have a coefficient of kinetic friction of 0.15.
How far does the crate slide on the horizontal floor if it continues to
experience a friction force.
1 1 1 1
− fd + Wotherforces = ( mv 2f + mgy f + kx 2f ) − ( mvi2 + mgyi + kxi2 )
2 2 2 2
1
−  k Nx + 0 = (0 + 0 + 0) − ( mvi2 + 0 + 0)
2
 k = 0.15, vi = 2.7m / s, N = ?
N − mg = 0
1
−  k mgx = − mvi2
2
2
v
x = i = 2.5m
2 k g
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 Conservation of Energy
❑ Energy is conserved
◼ This means that energy cannot be created nor
destroyed
◼ If the total amount of energy in a system
changes, it can only be due to the fact that
energy has crossed the boundary of the
system by some method of energy transfer

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 Ways to Transfer Energy
Into or Out of A System
❑ Work – transfers by applying a force and causing a
displacement of the point of application of the force
❑ Mechanical Waves – allow a disturbance to propagate
through a medium
❑ Heat – is driven by a temperature difference between
two regions in space
❑ Matter Transfer – matter physically crosses the
boundary of the system, carrying energy with it
❑ Electrical Transmission – transfer is by electric
current
❑ Electromagnetic Radiation – energy is transferred by
electromagnetic waves
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Energy of Various Objects and Phenomena

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 Ex 4: Connected Blocks in
Motion
❑ Two blocks are connected by a light string that passes over a
frictionless pulley. The block of mass m1 lies on a horizontal surface
and is connected to a spring of force constant k. The system is
released from rest when the spring is unstretched. If the hanging
block of mass m2 falls a distance h before coming to rest, calculate the
coefficient of kinetic friction between the block of mass m1 and the
surface. − fd + W
 = KE + PE
otherforces

1
PE = PE g + PE s = (0 − m2 gh) + ( kx 2 − 0)
2
1
−  k Nx + 0 = −m2 gh + kx 2
2
N = mg and x=h 1
m2 g − kh
1
−  k m1 gh = −m2 gh + kh 2 k = 2
2 m1 g
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 Power
❑ Work does not depend on time interval
❑ The rate at which energy is transferred is
important in the design and use of practical
device
❑ The time rate of energy transfer is called power
❑ The average power is given by
W
P=
t
◼ when the method of energy transfer is work

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 Instantaneous Power
❑ Power is the time rate of energy transfer. Power
is valid for any means of energy transfer
❑ Other expression W Fx
P= = = Fv
t t
❑ A more general definition of instantaneous
power   
W dW dr 
P = lim = =F = F v
t →0 t dt dt
 
P = F  v = Fv cos 
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 Units of Power
❑ The SI unit of power is called the watt
◼ 1 watt = 1 joule / second = 1 kg . m2 / s3
❑A unit of power in the US Customary
system is horsepower
◼ 1 hp = 550 ft . lb/s = 746 W
❑ Units
of power can also be used to
express units of work or energy
◼ 1 kWh = (1000 W)(3600 s) = 3.6 x106 J

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 Power Delivered by an Elevator
Motor
❑ A 1000-kg elevator carries a maximum load of 800 kg. A
constant frictional force of 4000 N retards its motion upward.
What minimum power must the motor deliver to lift the fully
loaded elevator at a constant speed of 3 m/s?
Fnet , y = ma y

T − f − Mg = 0
T = f + Mg = 2.16 104 N
P = Fv = (2.16 104 N )(3m / s)
= 6.48 104W
P = 64.8kW = 86.9hp
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