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Set 3 Problem number 28
By how much does the elastic PE
of a system change if an object of mass 7.6 kg and moving at 3.7 m/s is caught by and
stopped by a rubber band?
- In stretching the rubber band's
thermal energy increases by 19.24 Joules; this thermal energy is dissipated into the
surrounding air.
- If the object is then accelerated
by the rubber band as it snaps back, what will be its velocity at the instant the rubber
band goes slack?
An object of mass 7.6 kg moving at
3.7 m/s has kinetic energy .5 m v^2 = .5 * 7.6 kg * ( 3.7 m/s)^2 = 52.022 J.
- After being stopped by the rubber
band its KE is 0 and the PE of the rubber band is now 52.022 J.
- After the thermal energy of the
rubber band has been dissipated into the surrounding air is PE will be 52.022 J - 19.24 J =
32.782 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 * 32.782 J / 7.6 kg) = 2.93 m/s.
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