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Re: oscillations test question



You streatched it additionally by increasing the g field?

How about "doing it" horizontally using kinesthetic energy?

bc

Robert Cohen wrote:

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-----
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=
n
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?
=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 =
in
my mind is zero. By definition, a system at rest in stable
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=
y
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.

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.