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Law of the Conservation of
Mechanical Energy
When gravity is the only external influence on
an object, its mechanical energy remains
constant.
• FORMULA: (PE + KE) = C where C is
some constant value
Law of the Conservation of
Mechanical Energy
EXAMPLE: When a ball is thrown upward it slows
down, losing kinetic energy. However, as height
increases the value of potential energy also
increases. As it begins to fall back down its
velocity increases, increasing its kinetic energy
while its potential energy is reduced as it moves to
a lower height.
In other word, as kinetic energy decreases, potential
energy increases and vice-versa, keeping the sum of the
energies constant.
Principle of Work and Energy
• The work produced by a force is equal to
the change in energy that it produces in an
object on which it acts.
• FORMULA: W = ∆KE + ∆PE + ∆TE
where KE is kinetic energy, PE is potential
energy, and TE is thermal energy (heat,
which is often a product of a mechanical
process)
Principle of Work and Energy -
Application
• GIVEN: A ball with a mass of 2 kg is
dropped from a height of 1.5 m
• FIND: Its velocity immediately prior to
impact
Principle of Work and Energy
Application (continued)
Total mechanical energy possessed by the
ball:
PE + KE = C
(wt * h) + ½(mv2
) = C
wt = (2 kg)(9.81 m/s2
) = 19.62 N
Principle of Work and Energy
Application (continued)
At the start of this event, the ball is held so
that 100% of the energy is potential energy
and KE = 0
(19.62 N)(1.5 m) + 0 = C
29.43 J = C
Principle of Work and Energy
Application (continued)
Immediately before impact the height is 0 so
therefore PE = 0
(Remember, the sum of PE + KE stays the
same)
PE + KE = C = 29.43 J
0 + ½(mv2
) = 29.43 J
½(2 kg)(v2
) = 29.43 J
Principle of Work and Energy
Application (continued)
v2
= 29.43 J/kg
1 J = 1 Nm, 29.43 N = (1 kg)(29.43 m/s2
),
29.43 N/kg = 29.43 m/s2
Therefore 29.43 J/kg = (29.43 Nm)/(1 kg) =
(29.43 N/kg)(1m) = 29.43 m2
/s2
v2
= 29.43 m2
/s2
v = 5.42 m/s
WORK SAMPLE PROBLEM
ON PAGE 408
Practice Problems:
pp 411- 412
Introductory Problems 9 and 10
Additional Problems 7, 8, and 9

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426 45 conservation of mechanical energy

  • 1. Law of the Conservation of Mechanical Energy When gravity is the only external influence on an object, its mechanical energy remains constant. • FORMULA: (PE + KE) = C where C is some constant value
  • 2. Law of the Conservation of Mechanical Energy EXAMPLE: When a ball is thrown upward it slows down, losing kinetic energy. However, as height increases the value of potential energy also increases. As it begins to fall back down its velocity increases, increasing its kinetic energy while its potential energy is reduced as it moves to a lower height. In other word, as kinetic energy decreases, potential energy increases and vice-versa, keeping the sum of the energies constant.
  • 3. Principle of Work and Energy • The work produced by a force is equal to the change in energy that it produces in an object on which it acts. • FORMULA: W = ∆KE + ∆PE + ∆TE where KE is kinetic energy, PE is potential energy, and TE is thermal energy (heat, which is often a product of a mechanical process)
  • 4. Principle of Work and Energy - Application • GIVEN: A ball with a mass of 2 kg is dropped from a height of 1.5 m • FIND: Its velocity immediately prior to impact
  • 5. Principle of Work and Energy Application (continued) Total mechanical energy possessed by the ball: PE + KE = C (wt * h) + ½(mv2 ) = C wt = (2 kg)(9.81 m/s2 ) = 19.62 N
  • 6. Principle of Work and Energy Application (continued) At the start of this event, the ball is held so that 100% of the energy is potential energy and KE = 0 (19.62 N)(1.5 m) + 0 = C 29.43 J = C
  • 7. Principle of Work and Energy Application (continued) Immediately before impact the height is 0 so therefore PE = 0 (Remember, the sum of PE + KE stays the same) PE + KE = C = 29.43 J 0 + ½(mv2 ) = 29.43 J ½(2 kg)(v2 ) = 29.43 J
  • 8. Principle of Work and Energy Application (continued) v2 = 29.43 J/kg 1 J = 1 Nm, 29.43 N = (1 kg)(29.43 m/s2 ), 29.43 N/kg = 29.43 m/s2 Therefore 29.43 J/kg = (29.43 Nm)/(1 kg) = (29.43 N/kg)(1m) = 29.43 m2 /s2 v2 = 29.43 m2 /s2 v = 5.42 m/s
  • 10. Practice Problems: pp 411- 412 Introductory Problems 9 and 10 Additional Problems 7, 8, and 9