426 45 conservation of mechanical energy
10
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
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Transcript of 426 45 conservation of mechanical energy
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- 412Introductory Problems 9 and 10
Additional Problems 7, 8, and 9
