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Law of the Conservation of Mechanical Energy

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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

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.

- 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 = DKE + DPE + DTE where KE is kinetic energy, PE is potential energy, and TE is thermal energy (heat, which is often a product of a mechanical process)

- 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

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

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

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

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

pp 411- 412

Introductory Problems 9 and 10

Additional Problems 7, 8, and 9