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Set 3 Problem number 28
By how much does the elastic PE
of a system change if a rubber band snags and brings to rest an object of mass 8.3 kg
moving at 4.3 m/s?
- As it stretches the thermal energy
of the rubber band increases by 28.39 Joules.
- Before snapping back and sending
the object on its way, this excess thermal energy is dissipated into the surrounding air.
- What velocity will the object then
attain as the rubber band snaps back?
An object of mass 8.3 kg moving at
4.3 m/s has kinetic energy .5 m v^2 = .5 * 8.3 kg * ( 4.3 m/s)^2 = 76.73351 J.
- After being stopped by the rubber
band its KE is 0 and the PE of the rubber band is now 76.73351 J.
- After the thermal energy of the
rubber band has been dissipated into the surrounding air is PE will be 76.73351 J - 28.39 J =
48.34351 J.
- If this potential energy is then
recovered by the object in the form of KE its velocity will be v = `sqrt(2 KE / m) =
`sqrt( 2 * 48.34351 J / 8.3 kg) = 3.41 m/s.
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