equilibrium position
of each spring. Weigh the small
weight on the scale and record its mass in kilograms. Why does the mass have to
be in kilograms?
Attach the weight to each spring one at a time, and use the ruler to measure
the
displacement
. An easy way to do this is to measure the length of the spring,
and then subtract the equilibrium length.
Calculate the gravitational force exerted by the mass on the spring.
F
g
= mg
Where F
g
is the gravitational force, in Newtons, m is the mass of the weight, in
kilograms, and g is the gravitational constant of Earth, equal to 9.81 m/s
2
.
Set the gravitational force (F
g
) equal to the force exerted by the spring (F). Why
can you make these two variables equivalent? Use Hooke’s law to calculate the
spring constant, k, for each spring.
Results:
Springs with larger spring constants will have smaller displacements than springs
with lesser spring constants for the same mass added.
Why?
Hooke’s Law is a representation of
linear elastic
deformation. Elastic
means that the spring will return to its original form once the outside force (the
mass) is removed. Linear describes the relationship between the force and the
displacement. The fact that the spring constant is a constant (it is a property of the
spring itself), shows that the relationship is linear.
Of course,
Hooke’s Law
only remains true when the material is elastic. If
a spring is permanently deformed (by something like crushing or overstretching),
it will no longer return to its original position. If you have ever played with a
slinky and accidentally stretch it too far or bent it out of shape, you’ll know that
it doesn’t perform like it is supposed to afterward. For
Hooke’s Law
to work
properly, the parts of the equation have to be in the correct units. Without
consistent units, the equation is meaningless.
You can set the gravitational force exerted by the mass on the spring equal to the
force exerted by the spring due to
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