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A related question...
If the 200 grams is then stretched a distance x from the new equilibr=
ium point, the potential energy relative to the new equilibrium point=
is equal to 1/2 kx^2. What do you call the quantity 1/2 kx^2? The =
difference in the elastic-gravitational potential energy?
____________________________________________
Robert Cohen; rcohen@po-box.esu.edu; 570-422-3428; http://www.esu.edu=
/~bbq
Physics, East Stroudsburg Univ., E. Stroudsburg, PA 18301
-----Original Message-----n
From: Carl E. Mungan [mailto:mungan@USNA.EDU]
Sent: Wednesday, November 20, 2002 2:16 PM
=20
I'm interested in hearing what folks think of the following questio=
in the test bank for Giancoli:.
=20
Chap 14 #11. Two hundred grams hung on a spring stretches it 8.4 cm=
How much energy is stored when stretched 8.4 cm?in
=20
Before reading the rest, decide how you would answer this and why.
=20
While a number of answers are possible, the most reasonable answer =
my mind is zero. By definition, a system at rest in stabley
equilibrium is at a potential minimum.
=20
The purported answer of 82 mJ only considers the increase in the
elastic potential energy and neglects the decrease in the
gravitational potential energy.
=20
It is true that no zero points are specified. This is why I can onl=
say my answer is reasonable, and other interpretations are possible=.
Comments?
This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.