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

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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).

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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)
= J

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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 =
v = 125 m/s

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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

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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

7. Elastic Collision For all collisions, momentum is conserved.
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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

8. Is this collision elastic?
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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 = -45,000J

9. 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

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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

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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 = m2/s2
vf = 2.98 m/s

12. W = DKE Þ F*Dx = 1/2m(vf2 – vo2) 4.50Dx = 1/2(0.105)(3.82 – 1.32)
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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 = *(12.75)
4.50Dx = J
Dx = m

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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 = J
F = N