1. Conservation of Energy
November 18,

2. The conservation of energy
In a closed system, energy is neither created nor destroyed. Energy simply changes from one form to another.
Einitial = Efinal

3. 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

4. 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

5. 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 with friction / air resistance to say:
Einitial – Efriction = Efinal
Conservation of energy is a very useful tool to solve physics problems!
Many problems that can be solved using Newton’s laws can be solved much easier by applying conservation of energy.

6. h=0 N mg
300
h=0.5m
v=?
u=0
Determine the final velocity of a block that slides down a frictionless ramp. d
Method 1: Newton’s law
What steps would you have to take?

7. h=0 N mg
300 B A
h=0.5m
v=?
u=0
Determine the final velocity of a block that slides down a frictionless ramp. d
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

8. h=0 N mg
300 B A
h=0.5m
v=?
u=0
Determine the final velocity of a block that slides down a frictionless ramp. d
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
vf2 = vi2 + 2ad = 2gsinθ
v = 3.2 m/s

9. Method 2: Energy conservationmg
300 B A
h=0.5m
v=?
u=0
Determine the final velocity of a block that slides down a frictionless ramp. d
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
As the object slides down the plane its PE becomes transformed into KE.
vf2 = vi2 + 2ad = 2gsinθ
v = 3.2 m/s

10. Method 2: Energy conservationmg
300
h=0.5m
v=?
u=0
Determine the final velocity of a block that slides down a frictionless ramp. d
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
As the object slides down the plane its PE becomes transformed into KE.
PEinitial = KEfinal
mgh = ½ mv2
vf2 = vi2 + 2ax = 2gsinθ
v = 3.2 m/s

11. Method 2: Energy conservationmg
300
h=0.5m
v=?
u=0
Determine the final velocity of a block that slides down a frictionless ramp. d
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
As the object slides down the plane its PE becomes transformed into KE.
PEinitial = KEfinal
mgh = ½ mv2
v = 3.2 m/s
vf2 = vi2 + 2ad = 2gsinθ
We get the same answer, but the energy conservation answer is much easier.
v = 3.2 m/s

12. 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

13. 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.
m cancels on both sides
mgh = ½ mv2
a baby, an elephant and you would reach the bottom at the same speed !!!!!

14. Other examples of conservation of energy: Up and down the trackB 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

15. Other examples of conservation of energy: Up and down the trackB 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

16. Other examples of conservation of energy: Up and down the trackB 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

17. Other examples of conservation of energy: Up and down the trackB 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

18. Other examples of conservation of energy: Up and down the trackB C
At what point does the ball have the greatest speed?
1. A
2. B
3. C
4. D

19. 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
EXAMPLE 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 !!!!!

20. A pendulum swings back and forth
A pendulum swings back and forth. At which point or points along the pendulum’s path …
Is PE greatest?
Is KE greatest?
Is speed greatest?
G & A D
EXAMPLE 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.

21. 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
What is our strategy? v1
40m
20m

22. 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 v1
mgh1 + ½ mv12 = mgh2 + ½ mv22
40m
20m

23. 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 v1
mgh1 + ½ mv12 = mgh2 + ½ mv22
m cancels out ; v2 = 0
gh1 + ½ v12 = gh2
40m
20m
v12 = 2g( h2 – h1)
v1 = 22 m/s

24. 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:
1.85 m A B

25. 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:
1.85 m A B
PEA + KEA = PEB + KEB

26. 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 b. constant friction force of 16 N acts
on the object as it slides down.
What happens to energy in each case?

27. on the object as it slides down.
● 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.
all PE was transformed into KE 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.

28. on the object as it slides down.
● 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.
all PE was transformed into KE 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 = final energy initial energy – Ffr d = final energy
Wfr = – Ffr d decreases KE of the object

29. on the object as it slides down.
● 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.
all PE was transformed into KE 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 = final energy initial energy – Ffr d = final energy
Wfr = – Ffr d decreases KE of the object
PEA = KEB PEA – Ffr d = KEB

30. on the object as it slides down.
● 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.
all PE was transformed into KE 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 = final energy initial energy – Ffr d = final energy
Wfr = – Ffr d decreases KE of the object
PEA = KEB PEA – Ffr d = KEB
mgh = ½ mv mgh – Ffr d = ½ mv2
20 – 16 = 2.5 v2
v = 1.4 m/s