1. Conservation of Energy
Physics

2. Types of Energies Conservative Energies Non-Conservative Energies
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Types of Energies
Conservative Energies
Kinetic
Gravitational Potential
Spring Potential
Rotational
Non-Conservative Energies
Chemical
Heat
Sound
Friction
Thrust
Mechanical Energy
Dissipated

3. Conservation of Energy
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Conservation of Energy
In a closed system, energy will transfer from one type to another without the total energy increasing or decreasing.
What determines if energy is changing?
Money
Energy
Checking
Kinetic
Changing Speed
Potential (Grav. & Spring)
Changing Height or Stretch/Compression
Savings
Spend Money
Dissipated
Friction, Sound, Heat
Force with Displacement
Gain Money
Work

4. Ball Rolling Down a Ramp without Air Friction
PEg KE Diss

5. Ball Rolling Down a Ramp without Air Friction
PEg KE Diss

6. Ball Rolling Down a Ramp with Air Friction
PEg KE Diss

7. Ball Rolling Down a Ramp with Air Friction
PEg KE Diss

8. Book Falling without Air Friction
PEg KE Diss

9. Book Falling without Air Friction
PEg KE Diss

10. Ball Bouncing with Air Friction
PEg KE Diss PEs

11. Ball Bouncing with Air Friction
PEg KE Diss PEs

12. Conservation of Energy
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Conservation of Energy
∆𝑬=𝟎
∆𝑲𝑬
∆𝑷𝑬 𝒈
∆𝑷𝑬 𝒔
𝑫𝒊𝒔𝒔𝒊𝒑𝒂𝒕𝒆𝒅
∆𝑲𝑬+ ∆𝑷𝑬 𝒈 + ∆𝑷𝑬 𝒔 +𝑫𝒊𝒔𝒔.=𝟎
∆𝑲𝑬= 𝟏 𝟐 𝒎 𝒗 𝒇 𝟐 − 𝒗 𝒐 𝟐
∆𝑷𝑬 𝒈 =𝒎𝒈 𝒉 𝒇 − 𝒉 𝒐
∆𝑷𝑬 𝒔 = 𝟏 𝟐 𝒌 𝒙 𝒇 𝟐 − 𝒙 𝒐 𝟐

13. Example #1 0 = DKE + DPEg 0 = 1/2m(vf2-vo2) + mgDh
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Example #1
A 2.5-kg mass is dropped from a 53 m cliff. Using the Law of Conservation of Energy, how fast will the mass be moving right before it hits the bottom of the cliff?
0 = DKE + DPEg
0 = 1/2m(vf2-vo2) + mgDh

14. The mass will be moving at 32 m/s
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0 = 1/2m(vf2-vo2) + mgDh
0 = 1/2(2.5)(vf2–02)+(2.5)(9.8)(-53)
0 = 1.25*vf
1.25*vf2 =
vf2 =
vf = 32.2 m/s
The mass will be moving at 32 m/s

15. Example #2 0 = DKE + DPEg + Diss 0 = 1/2m(vf2-vo2) + mgDh + Diss
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Example #2
A 3.0-kg mass is dropped from a 62 m cliff and 700 J of energy is dissipated. How fast will the mass be moving right before it hits the bottom of the cliff?
0 = DKE + DPEg + Diss
0 = 1/2m(vf2-vo2) + mgDh + Diss

16. 0 = 1/2m(vf2-vo2) + mgDh + Diss
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0 = 1/2m(vf2-vo2) + mgDh + Diss
0 = 1/2(3)(vf2–02)+(3)(9.8)(-62)+700
0 = 1.5*vf
0 = 1.5*vf
1.5*vf2 =
vf2 = 748.5
vf = 27.4 m/s
The mass will be moving at 27 m/s

17. Example #3 0 = DKE + DPEs 0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2)
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Example #3
A 2.5-kg mass moving horizontally at 3.6 m/s hits a spring with a stiffness of 200 N/m. Using the Law of Conservation of Energy, how far will the spring compress when the mass comes to a rest?
0 = DKE + DPEs
0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2)

18. 0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2)
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0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2)
0 = 1/2(2.5)(02–3.62)+1/2(200)(xf2–02)
0 = *xf2
100*xf2 = 16.2
xf2 = 0.162
xf = m
The spring will compress 40.2 cm

19. Example #4 0 = DKE + DPEs 0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2)
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Example #4
A 50-g mass is being pushed by a spring with a k=150 N/m and compressed to 25 cm.
The spring is then released. How fast will the mass be moving after the spring has fully uncompressed?
0 = DKE + DPEs
0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2)

20. 0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2)
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0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2)
0 = 1/2(0.050)(vf2 – 02)
+1/2(150)(02 – 0.252)
0 = 0.025*vf2 –
0.025*vf2 =
vf2 = 187.5
vf = 13.7 m/s
The mass will be moving 13.7 m/s