6.3. Energy Conservation http://www.ck12.org
Notice that the equation in part (c),
1
2 mv
(^2) =mgh, can be solved asv (^2) = 2 gh, and is identical to the kinematic equation:
vf^2 =vi^2 + 2 a∆x, wherevf=0, anda=−g.
http://demonstrations.wolfram.com/PotentialAndKineticEnergiesOfAFallingObject/
Illustrative Example 2
Qualitatively describe the kinetic and potential energies for the bob on the pendulum inFigure6.19 at positions
P 1 ,P 2 ,P 3.
FIGURE 6.19
Answers
a.P 1 : The bob is momentarily at rest at its highest position, so all the energy is in the form of gravitational potential
energy.
b.P 2 : The bob is at its lowest position, so all of the energy is in the form of kinetic energy.
c.P 3 : The bob is between its lowest and highest positions, so it has a combination of kinetic and potential energies.
Illustrative Example 3
The kinetic and potential energies used to solve problems related to the gravitational field of the earth can often be
reduced to kinematic equations. Why then, you might wonder, do we bother with energy in the first place? Perhaps
the next example will clarify this.
A sled slides from rest from a height of 4.00 m down a frictionless curved ramp. What is the maximum speed that
the sled can achieve? SeeFigure6.20
Answer:The sled begins its ride with all potential energy and has its maximum speed when all of its potential energy
has been transformed to kinetic energy. This occurs when the sled comes off the ramp, thus:Ei=Ef,KEi+PEi=
KEf+PEf: 0+mgh=^12 mv^2 + 0
mgh=1
2
mv^2 ,gh=1
2
v^2 ,v^2 = 2 gh,v=√
2 ghv=√
2 ( 9. 81 )( 4. 00 )
v= 8. 86 m/s