Chapter 8 Conservation of Energy EXAMPLES

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Chapter 8Conservation of Energy EXAMPLES
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Example 8.1 Free Fall(Example 8.1 Text book)

• Determine the speed of the ball at y above the
  ground
• The sum K U remains constant
• At h Ui mgh Ki 0
• At y Kf ½mvf2 Uf mgy
• In general Conservation of Energy
• Ki Ui Kf Uf ?
• 0 mgh ½mvf2 mgy ?
• Solving for vf
• vf is independent of the mass !!!

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Example 8.2 Roller Coaster Speed

• Using Mechanical Energy Conservation
  (Frictionless!)
• ½mvf2 mgyf ½mvi2 mgyi
• Only vertical differences matter!
• Horizontal distance doesnt matter!
• Mass cancels!
• Find Speed at bottom?
• Known yi y 40m, vi 0,
• yf 0, vf ? ?
• 0 mgyi ½mvf2 0 ?
• vf2 2gyi 784m2/s2 ?
• v2 28 m/s

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Example 8.3 Spring-Loaded Gun (Example 8.3
Text book)

• Choose point A as the initial point and C as the
  final point
• (A). Find the Spring Constant k ?
• Known vA 0, yA 0 , xA yB 0.120m
• vC 0 , yC 20m, UC mgyC , m 35.0g
• EC EA ?
• KC UgC UsC KA UgA UsA
• ½mvC2 mgyC ½kxC2 ½mvA2 mgyA ½kxA2
• ? 0 mgyC 0 0 0 ½kxA2
• ? ½kxA2 mgyC ? k 2mgyc/xA2
• 2(0.0350kg)(9.80m/s2)(20.0m)/(0.120m)2
• k 953 N/m

yC
yB
yA
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Example 8.3 Spring-Loaded Gun, final

• (B). Find vB ?
• Use EB EA
• KB UgB USB KA UgA USA
• ½mvB2 mgyB ½kxB2
• ½mvA2 mgyA ½kxA2 ?
• ½mvB2 mgyB 0 0 0 ½kxA2
• vB2 (kxA2 2mgyB)/m
• vB2 388.1m2/s2 ?
• vB 19.7 m/s

yC
yB
yA
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Example 8.4 Ramp with Friction(Example 8.7 Text
book)

• Problem the 3.0 kg crate slides down the rough
  ramp. If vi 0, yi 0.5m , yf 0
  ƒk 5N
• (A). Find speed at bottom vf
• At the top Ei Ki Ugi 0 mgyi
• At the bottom Ef Kf Ugf ½ m vf2 0
• Recall If friction acts within an isolated
  system
• ?Emech ?K ?U Ef Ei ƒk d ?
• ½ m vf2 mgyi ƒk d
• Solve for vf

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Example 8.4 Ramp with Friction, final

• (B). How far does the crate slide on the
  horizontal floor if it continues to experience
  the same friction force ƒk 5N
• The total ?Emech is only kinetic (the potential
  energy of the system remains fixed)
• ?Emech ?K Kf Ki ƒk d
• Where Kf ½ m vf2 0
• Ki ½ m vi2 9.68 J
• Then Kf Ki 0 9.68 J (5N)d ?
• d (9.68/5) 1.94 m

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Example 8.5 Motion on a Curve Track
(Frictionless)

• A child of mass m 20 kg starts sliding from
  rest. Frictionless!
• Find speed v at the bottom.
• ?Emech ?K ?U
• ?Emech (Kf Ki) (Uf Ui) 0
• (½mvf2 0) (0 mgh) 0
• ½mvf2 mgh 0 ?
• Same result as the child is falling vertically
  trough a distance h!

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Example 8.5 Motion on a Curve Track (Friction)

• If a kinetic friction of ƒk 2N acts on the
  child and the length of the curve track is 50 m,
  find speed v at the bottom.
• If friction acts within an isolated system
• ?Emech ?K ?U ƒk d
• ?Emech (Kf Ki) (Uf Ui) ƒk d
• ?Emech ½mvf2 mgh ƒk d
• ?Emech ½(20) vf2 20(10)(2) 100J

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Example 8.6 Spring-Mass Collision(Example 8.8
Text book)

• Frictionless!
• K Us Emech remains constant
• (A). Assuming m 0.80kg vA 1.2m/s
  k 50N/m
• Find maximum compression of the spring after
  collision (xmax)
• EC EA ? KC UsC KA UsA
• ½mvC2 ½kxmax2 ½mvA2 ½kxA2
• 0 ½kxmax2 ½mvA2 0 ?

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Example 8.6 Spring-Mass Collision, final

• (B). If friction is present, the energy decreases
  by
• ?Emech ƒkd
• Assuming ?k 0.50 m 0.80kg vA 1.2m/s k
  50N/m
• Find maximum compression of the spring after
  collision xC
• ?Emech ƒk xC ?knxC ?kmgxC ?
• ?Emech 3.92xC (1)
• Using ?Emech Ef Ei
• ?Emech (Kf Uf) (Ki Ui)
• ?Emech 0 ½kxC2 ½mvA2 0
• ?Emech 25xC2 0.576 (2)
• Taking (1) (2) 25xC2 0.576 3.92xC
• Solving the quadratic equation for xC
• xC 0.092m lt 0.15m (frictionless)
• Expected! Since friction retards the motion of
  the system
• xC 0.25m Does not apply since the mass must
  be to the right of the origin.

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Example 8.7 Connected Blocks in Motion(Example
8.9 Text book)

• The system consists of the two blocks, the
  spring, and Earth. Gravitational and potential
  energies are involved
• System is released from rest when spring is
  unstretched.
• Mass m2 falls a distance h before coming to rest.
  
• The kinetic energy is zero if our initial and
  final configurations are at rest ?K 0
• Find ?k

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Example 8.7 Connected Blocks in Motion, final

• Block 2 undergoes a change in gravitational
  potential energy
• The spring undergoes a change in elastic
  potential energy
• ?Emech ?K ?Ug ?US ?Ug ?US Ugf
  Ugf Usf Usi
• ?Emech 0 m2gh ½kh2 0
• ?Emech m2gh ½kh2 (1)
• If friction is present, the energy decreases by
  
• ?Emech ƒkh ?km1gh (2)
• Taking (1) (2) m2gh ½kh2 ?k m1gh
• ?k m1gh m2gh ½kh2 ?