### click - Uplift Education

```Conservation of Energy
November 21 2014
The conservation of energy
 In a closed system, energy is neither created nor destroyed.
Energy simply changes from one form to another.
Einitial = Efinal
The conservation of energy
 In a closed system, energy is neither created nor destroyed.
Energy simply changes from one form to another.
Einitial = Efinal
 Mostly, we will use this in situations without friction to say:
PEinitial + KEinitial = PEfinal + KEfinal
The conservation of energy
 In a closed system, energy is neither created nor destroyed.
Energy simply changes from one form to another.
Einitial = Efinal
 Mostly, we will use this in situations without friction to say:
PEinitial + KEinitial = PEfinal + KEfinal
 We can also use this in situations where energy is
converted to heat due to friction or air resistance to say:
Einitial – Eheat = Efinal
The conservation of energy
 In a closed system, energy is neither created nor destroyed.
Energy simply changes from one form to another.
Conservation
of energy is a very
Einitial = Efinal
useful tool to solve physics
 Mostly,problems!
we will use this in situations without friction to say:
PEinitial + KEinitial = PEfinal + KEfinal
Many problems that can be solved
Newton’s
laws can
befriction
solved/ air
 We canusing
also use
this in situations
with
resistance
say: by applying
muchto easier
Einitial – Efriction
= Efinal
conservation
of energy.
h
d=
sinθ
v=?
h=0
N
d
u=0
h=0.5m
300
Method 1: Newton’s law
What steps would you have to
take?
Determine the final
velocity of a block that
slides down a
frictionless ramp.
h
d=
sinθ
v=?
h=0 B
N
d
u=0
A
h=0.5m
300
Method 1: Newton’s law
What steps would you have to
take?
Step 1: Draw FBD
Step 2: Write Fnet equation and
solve for a
Step 3: Use motion equations to
solve for v
Determine the final
velocity of a block that
slides down a
frictionless ramp.
h
d=
sinθ
v=?
h=0 B
N
d
u=0
A
h=0.5m
300
Method 1: Newton’s law
What steps would you have to
take?
Step 1: Draw FBD
Step 2: Write Fnet equation and
solve for a
Step 3: Use motion equations to
solve for v
F
mgsinθ
a=
=
= gsinθ
m
m
 h 
2
2
vf = vi + 2ad = 2gsinθ 

 sinθ 
v = 3.2 m/s
v = 2gh
Determine the final
velocity of a block that
slides down a
frictionless ramp.
h
d=
sinθ
v=?
h=0 B
N
d
u=0
A
h=0.5m
300
Determine the final
velocity of a block that
slides down a
frictionless ramp.
Method 1: Newton’s law
Method 2: Energy conservation
What steps would you have to
take?
Step 1: Draw FBD
Step 2: Write Fnet equation and
solve for a
Step 3: Use motion equations to
solve for v
F
mgsinθ
a=
=
= gsinθ
m
m
 h 
2
2
vf = vi + 2ad = 2gsinθ 

 sinθ 
v = 3.2 m/s
v = 2gh
As the object slides down the plane its
PE becomes transformed into KE.
h
d=
sinθ
v=?
h=0
N
d
u=0
h=0.5m
300
Determine the final
velocity of a block that
slides down a
frictionless ramp.
Method 1: Newton’s law
Method 2: Energy conservation
What steps would you have to
take?
Step 1: Draw FBD
Step 2: Write Fnet equation and
solve for a
Step 3: Use motion equations to
solve for v
F
mgsinθ
a=
=
= gsinθ
m
m
 h 
2
2
vf = vi + 2ax = 2gsinθ 

 sinθ 
v = 3.2 m/s
v = 2gh
As the object slides down the plane its
PE becomes transformed into KE.
PEinitial = KEfinal
mgh = ½ mv2
h
d=
sinθ
v=?
h=0
N
d
u=0
h=0.5m
300
Determine the final
velocity of a block that
slides down a
frictionless ramp.
Method 1: Newton’s law
Method 2: Energy conservation
What steps would you have to
take?
Step 1: Draw FBD
Step 2: Write Fnet equation and
solve for a
Step 3: Use motion equations to
solve for v
F
mgsinθ
a=
=
= gsinθ
m
m
 h 
2
2
vf = vi + 2ad = 2gsinθ 

 sinθ 
v = 3.2 m/s
v = 2gh
As the object slides down the plane its
PE becomes transformed into KE.
PEinitial = KEfinal
mgh = ½ mv2
v = 2gh
v = 3.2 m/s
We get the same answer, but the
much easier.
Examples of Energy Conservation
Free fall in the absence of air resistance.
A 10 kg ball is dropped from a height of 102 m.
What is its velocity when it hits the ground?
Initial E = mgh = 10kg * 9.8 m/s2 *102 m = 10000J
Final E = ½ mv2 = 10000J
v = √(2000) = 45 m/s
Free fall with air resistance.
A 10 kg ball is dropped from a height of 102 m.
When it hits the ground, it has a final velocity of
40m/s. How much energy was ‘lost’ due to air
resistance?
Initial E = mgh = 10kg * 9.8 m/s2 *102 m = 10000J
Final E = ½ mv2 = 0.5*10kg*(40m/s)2 = 8000 J
Eheat = Initial E – Final E = 10000J – 8000 J = 2000 J
You Do
A child of mass m is released from rest at
the top of a water slide, at height h = 8.5 m
above the bottom of the slide. Assuming
that the slide is frictionless because of the
water on it, find the child’s speed at the
bottom of the slide.
mgh = ½
mv2
v = 2gh = 13 m/s
m cancels
on both
sides
a baby, an elephant and you would reach
the bottom at the same speed !!!!!
Other examples of conservation of energy:
Up and down the track
A
B
C
Assuming that the track is frictionless, at what point(s) is the total energy
of the system the same as in point A ?
1. B
2. C
3. D
4. All of the above
5. none of the above
D
Other examples of conservation of energy:
Up and down the track
A
B
C
Assuming the track is frictionless, at what height will the ball end up?
1. same height as A
2. lower height than A
3. higher height than A
4. impossible to determine
D
Other examples of conservation of energy:
Up and down the track
A
B
C
If the track does have friction, at what height will the ball end up?
1. same height as A
2. lower height than A
3. higher height than A
4. impossible to determine
D
Other examples of conservation of energy:
Up and down the track
A
B
C
At what point(s) does the ball have a combination of PE and KE?
1. A
2. B
3. C
4. D
5. B and C
D
Other examples of conservation of energy:
Up and down the track
A
B
C
At what point does the ball have the greatest speed?
1. A
2. B
3. C
4. D
D
Three balls are thrown from the top of the cliff along paths A, B, and C with
the same initial speed (air resistance is negligible). Which ball strikes the
ground below with the greatest speed?
1. A
2. B
3. C
4. All strike with the same speed
h
Think about it for a minute … then when I say to, show
which answer is correct with a show of fingers
#4 is correct
All balls have the same initial energy, so all have the same KE when
they hit the ground.
It depends ONLY on HEIGHT and initial SPEED, not mass, not path !!!!!
A pendulum swings back and forth. At
which point or points along the
pendulum’s path …
1) Is PE greatest? G & A
2) Is KE greatest?
D
3) Is speed greatest? D
An ideal pendulum would keep going
forever. Why do real pendulums
eventually stop?
Because some energy is ‘lost’ due to
friction and air resistance.
We Do
A car (toy car – no engine) is at the top of a hill on a
frictionless track. What must the car’s speed be at the top of
the first hill if it can just make it to the top of the second hill?
v2 = 0
v1
40m
20m
What is our strategy?
We Do
A car (toy car – no engine) is at the top of a hill on a
frictionless track. What must the car’s speed be at the top of
the first hill if it can just make it to the top of the second hill?
v2 = 0
mgh1 + ½ mv12 = mgh2 + ½ mv22
v1
40m
20m
PE1 + KE1 = PE2 + KE2
We Do
A car (toy car – no engine) is at the top of a hill on a
frictionless track. What must the car’s speed be at the top of
the first hill if it can just make it to the top of the second hill?
v2 = 0
PE1 + KE1 = PE2 + KE2
mgh1 + ½ mv12 = mgh2 + ½ mv22
v1
m cancels out ; v2 = 0
gh1 + ½ v12 = gh2
40m
20m
v12 = 2g( h2 – h1)
v1 = 22 m/s
You Do
A simple pendulum consists of a 2.0 kg mass attached
to a string. It is released from rest at A
as shown. Its speed at the lowest point B is:
PEA + KEA = PEB + KEB
1.85 m
1
1
2
mv A + mghA = mv B 2 + mghB
2
2
1
0 + ghA = mv B 2 + 0
2
2ghA =
2×9.80m/s2 ×1.85m = 6m/s
A
B
vB =
You Do
A simple pendulum consists of a 2.0 kg mass attached
to a string. It is released from rest at A
as shown. Its speed at the lowest point B is:
A
1.85 m
B
● Example: An object of mass 4.0 kg slides 1.0 m down
an inclined plane starting from rest. Determine the
speed of the object when it reaches the bottom of
the plane if
a.
friction is neglected
b. constant friction force of 16 N acts
on the object as it slides down.
What happens to energy in each case?
● Example: An object of mass 4.0 kg slides 1.0 m down
an inclined plane starting from rest. Determine the
speed of the object when it reaches the bottom of
the plane if
a.
friction is neglected
all PE was transformed into KE
b. constant friction force of 16 N acts
on the object as it slides down.
Friction converts part of KE of the object
into heat energy. This energy equals to
the work done by the friction. We say that
the frictional force has dissipated energy.
● Example: An object of mass 4.0 kg slides 1.0 m down
an inclined plane starting from rest. Determine the
speed of the object when it reaches the bottom of
the plane if
a.
friction is neglected
all PE was transformed into KE
initial energy = final energy
b. constant friction force of 16 N acts
on the object as it slides down.
Friction converts part of KE of the object
into heat energy. This energy equals to
the work done by the friction. We say that
the frictional force has dissipated energy.
initial energy – Ffr d = final energy
Wfr = – Ffr d decreases KE of the object
● Example: An object of mass 4.0 kg slides 1.0 m down
an inclined plane starting from rest. Determine the
speed of the object when it reaches the bottom of
the plane if
a.
friction is neglected
all PE was transformed into KE
initial energy = final energy
b. constant friction force of 16 N acts
on the object as it slides down.
Friction converts part of KE of the object
into heat energy. This energy equals to
the work done by the friction. We say that
the frictional force has dissipated energy.
initial energy – Ffr d = final energy
Wfr = – Ffr d decreases KE of the object
PEA = KEB
PEA – Ffr d = KEB
● Example: An object of mass 4.0 kg slides 1.0 m down
an inclined plane starting from rest. Determine the
speed of the object when it reaches the bottom of
the plane if
a.
friction is neglected
all PE was transformed into KE
initial energy = final energy
b. constant friction force of 16 N acts
on the object as it slides down.
Friction converts part of KE of the object
into heat energy. This energy equals to
the work done by the friction. We say that
the frictional force has dissipated energy.
initial energy – Ffr d = final energy
Wfr = – Ffr d decreases KE of the object
PEA = KEB
mgh = ½ mv2
v = 2gh = 3.2 m/s
PEA – Ffr d = KEB
mgh – Ffr d = ½ mv2
20 – 16 = 2.5 v2
v = 1.4 m/s
```