AP Physics 1 (thanks to Gabe Kessler, OCHS!) Kinetic Energy AP Physics 1 (thanks to Gabe Kessler, OCHS!)

End Slide Kinetic Energy KE = 1/2mv2 DKE = KEf – KEo = 1/2mvf2 – 1/2mvo2 Kinetic Energy is the energy stored in the motion of an object. A change in Kinetic Energy will also mean a change in speed (magnitude of velocity).

End Slide A man with a mass of 88 kg is running at 18.0 m/s. What is his Kinetic Energy? KE = 1/2mv2 = 1/2(88)(18.02) = 14300 J

End Slide A plane has a mass of 180,000 kg and 1.41x108 J of Kinetic Energy. What is the plane’s speed? KE = 1/2mv2 1.41x108 = 1/2(180,000)(v2) 1.41x108 = 90,000v2 90,000 90,000 v2 = 15700 v = 125 m/s

End Slide A car of mass 3500 kg is moving at 20.0 m/s and slows to 5.0 m/s. What is DKE? DKE = 1/2m(vf2 – vo2) = 1/2(3500)(52 – 202) = 1750*(–375) = –656,250 J » –6.6x105 J

End Slide The same 3500-kg car is moving at 5.0 m/s. It stops and moves in reverse at -5.0 m/s. What is DKE? DKE = 1/2m(vf2 – vo2) = 1/2(3500)(-52 – 52) = 1750*(0) = 0 J

Elastic Collision For all collisions, momentum is conserved. End Slide Elastic Collision For all collisions, momentum is conserved. If Kinetic Energy is conserved (DKE = 0), then the collision is said to be elastic. If Kinetic Energy is not conserved (DKE ¹ 0), then the collision is inelastic. Realize that no collision is ever really perfectly elastic; however, some are close enough

Is this collision elastic? End Slide Is this collision elastic? Car 1 with a mass of 1200 kg moving at 25.0 m/s strikes Car 2 with half of Car 1’s mass while it’s sitting still. Car 2 flies off at 30.0 m/s while Car 1 moves at 10.0 m/s. DKE1 = 1/2m1(vf12 – vo12) = 1/2(1200)(102 – 252) = -315,000 J DKE2 = 1/2m2(vf22 – vo22) = 1/2(600)(302 – 02) = 270,000 J DKE1+DKE2 = -315,000+270,000 = -45,000J

Work-Energy Theorem The total work done on an open system will produce a change in kinetic energy. 𝑾= 𝟏 𝟐 𝒎 𝒗 𝒇 𝟐 − 𝟏 𝟐 𝒎 𝒗 𝒐 𝟐 𝑾 𝑻𝒐𝒕𝒂𝒍 =𝜮𝑭∗∆𝒙 𝜮𝑭=𝒎𝒂 𝑾=𝒎 𝒂∗∆𝒙 𝒗 𝒇 𝟐 = 𝒗 𝒐 𝟐 +𝟐𝒂∆𝒙 − 𝒗 𝒐 𝟐 − 𝒗 𝒐 𝟐 𝑾=𝒎 𝒗 𝒇 𝟐 − 𝒗 𝒐 𝟐 𝟐 𝒗 𝒇 𝟐 − 𝒗 𝒐 𝟐 =𝟐𝒂∆𝒙 𝟐 𝟐 𝒗 𝒇 𝟐 − 𝒗 𝒐 𝟐 𝟐 =𝒂∆𝒙 𝑾= 𝟏 𝟐 𝒎 𝒗 𝒇 𝟐 − 𝟏 𝟐 𝒎 𝒗 𝒐 𝟐

End Slide Work-Energy Theorem The total work done on an open system will produce a change in kinetic energy. 𝑾= 𝟏 𝟐 𝒎 𝒗 𝒇 𝟐 − 𝟏 𝟐 𝒎 𝒗 𝒐 𝟐 𝑲𝑬= 𝟏 𝟐 𝒎 𝒗 𝟐 𝑾= 𝑲𝑬 𝒇 − 𝑲𝑬 𝒐 𝑾=∆𝑲𝑬

End Slide A horizontal force of 3250 N is applied to a 110-kg mass initially at rest for a horizontal distance of 0.15 m. What is the final velocity of the mass? W = DKE F*Dx = 1/2m(vf2 – vo2) (3250)(0.15) = 1/2(110) (vf2 – 02) 487.5J = 55vf2 vf2 = 8.86 m2/s2 vf = 2.98 m/s

W = DKE Þ F*Dx = 1/2m(vf2 – vo2) 4.50Dx = 1/2(0.105)(3.82 – 1.32) End Slide A 105-g hockey puck is sliding across the ice at 1.3 m/s. A player exerts a constant 4.50-N force to give the puck a velocity of 3.8 m/s. For what distance did the player apply the force to the puck? W = DKE Þ F*Dx = 1/2m(vf2 – vo2) 4.50Dx = 1/2(0.105)(3.82 – 1.32) 4.50Dx = 0.0525*(12.75) 4.50Dx = 0.6694J Dx = 0.149 m

End Slide An ice skater with a mass of 52.0 kg is moving at 2.5 m/s and glides to a stop over a displacement of 24.0 m. What is the average force due to friction on the skater while she is stopping? W = DKE F*Dx = 1/2m(vf2 – vo2) F*(24.0)=1/2(52.0)(02 – 2.52) F*24.0 = -162.5J F = -6.77 N