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Potential energy is a slippery concept. To figure potential energy c=
orrectly, one must remember that there is no such thing as "the" pote=
ntial energy. There is only "a" potential energy. Potential energy i=
s always defined as the energy _difference_ between two points. Fo=
r example, the P.E. of a textbook held 2 m above the floor might be m=
g(1m) if there is a table between the book and the floor, and mg(2m) =
if there isn't. This is a subtle point that is often omitted for t=
he sake of clarity, thereby leading to confusion. Another subtle po=
int is that the change in energy is related to the work done.
The trouble with setting mgx =3D 1/2*k*x^2 is that the two sides of t=
he equation are not equal. The force on the mass is given by kx-mg. =
The work done (hence the change in energy) in moving the mass from t=
he top position to the equilibrium position is given by the negative =
of the integral of F*dx. This is the negative of the integral of (kx=
-mg)*dx. Integrating gives=20
W =3D mgx - 1/2*k*x^2. The work W isn't zero. In fact, it's precisel=
y 1/2 * k*x^2, which is the energy stored by simply stretching a spri=
ng, which is derived by integrating F=3D-kx. Substitution gives 1/2*=
k*x^2 =3D mgx - 1/2*k*x^2, which leads one right back to mg=3Dkx. =
=20
To find the mass, either use mg =3D kx, or get it from the period of =
oscillation of the mass-spring system.
Vickie Frohne=20
-----Original Message-----
=46rom: Forum for Physics Educators on behalf of Savinainen Antti
Sent: Sat 12/11/2004 2:16 PM
To: PHYS-L@LISTS.NAU.EDU
Subject: Spring question
=20
Hi,
a friend of mine asked a very simple question which could be formulat=
ed in the following way:
A vertical spring (spring constant k) is hanging freely.
A block of mass m is attached to the spring and the spring
streches. At the equilibrium position of the block, where it hangs mo=
tionless, the spring has streched by x. Determine the mass of the blo=
ck.
Solution 1:
At the equilibrium the net force on the block is zero
and hence kx =3D mg which gives m =3D kx/g.
Solution 2:
The change in gravitational potential energy of the block
is mgx. This energy must be in the mass-spring system and
equals 1/2kx*x (assuming no friction in the spring).
But this calculation gives m =3D 1/2* kx/g!!
Which solution is correct? What's wrong with the other
solution?
Regards,
Antti
Antti Savinainen, Ph.D.
Senior Lecturer in Physics and Mathematics
Kuopio Lyseo High School
Finland
E-mail: <antti.savinainen@kuopio.fi>
Website: <http://kotisivu.mtv3.fi/physics/>