#$&*
phy 201
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PE and KE
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is this the correct formula for solving for PE = .5 mv^2 = .5 kA^2
and KE?
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@& KE = 1/2 m v^2
or for rotation
KE = 1/2 I omega^2.
For KE, that's it.
The formula for PE depends on the nature of the force.
At position x the PE of an object which experiences a linear restoring force proportional to its displacement from equilibrium (i.e., for which F_elastic = - k x) is 1/2 k x^2.
You also have gravitational PE near the surface of the Earth, which can be written PE = m g y (PE with respect to the surface of the Earth), and for positions which are significantly further from the center of the Earth than the surface, PE = - G M m / r.*@