Conservation of Energy Physics

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

Conservation of Energy End Slide 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

Ball Rolling Down a Ramp without Air Friction PEg KE Diss

Ball Rolling Down a Ramp without Air Friction PEg KE Diss

Ball Rolling Down a Ramp with Air Friction PEg KE Diss

Ball Rolling Down a Ramp with Air Friction PEg KE Diss

Book Falling without Air Friction PEg KE Diss

Book Falling without Air Friction PEg KE Diss

Ball Bouncing with Air Friction PEg KE Diss PEs

Ball Bouncing with Air Friction PEg KE Diss PEs

Conservation of Energy End Slide 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

Example #1 0 = DKE + DPEg 0 = 1/2m(vf2-vo2) + mgDh End Slide 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

The mass will be moving at 32 m/s End Slide 0 = 1/2m(vf2-vo2) + mgDh 0 = 1/2(2.5)(vf2–02)+(2.5)(9.8)(-53) 0 = 1.25*vf2-1298.5 1.25*vf2 = 1298.5 vf2 = 1038.8 vf = 32.2 m/s The mass will be moving at 32 m/s

Example #2 0 = DKE + DPEg + Diss 0 = 1/2m(vf2-vo2) + mgDh + Diss End Slide 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

0 = 1/2m(vf2-vo2) + mgDh + Diss End Slide 0 = 1/2m(vf2-vo2) + mgDh + Diss 0 = 1/2(3)(vf2–02)+(3)(9.8)(-62)+700 0 = 1.5*vf2-1822.8+700 0 = 1.5*vf2-1122.8 1.5*vf2 = 1122.8 vf2 = 748.5 vf = 27.4 m/s The mass will be moving at 27 m/s

Example #3 0 = DKE + DPEs 0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2) End Slide 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)

0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2) End Slide 0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2) 0 = 1/2(2.5)(02–3.62)+1/2(200)(xf2–02) 0 = -16.2 + 100*xf2 100*xf2 = 16.2 xf2 = 0.162 xf = 0.402 m The spring will compress 40.2 cm

Example #4 0 = DKE + DPEs 0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2) End Slide 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)

0 = 1/2m(vf2-vo2)+1/2k(xf2-xo2) End Slide 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 – 4.6875 0.025*vf2 = 4.6875 vf2 = 187.5 vf = 13.7 m/s The mass will be moving 13.7 m/s