Hooke’s Law
 A mass at the end of a spring will displace the spring to a certain
displacement (x).
 The restoring force acts in springs to try and return the spring to its
un-stretched state
 This force opposes the displacement and follows Hooke’s Law: F = - kx
 The – sign can be omitted from problem solving as it only indicates
that the force direction is opposite to the displacement direction.
 This is not a constant force as force increases with x (so does the
acceleration too).
 The force of gravity is balanced by the spring force
Example 1)
A spring stretches 0.150 m when 0.300 kg hangs from it. Find the spring
constant.
When the mass hangs from the spring, the weight of the mass is equal to
the restoring force.
F = |kx| = |mg|
k = [(0.300 kg) (9.81 m/s2)]/[0.150 m]
= 19.6 N/m
Example 2)
The force constant of a spring is 48.0 N/m. It has a 0.25 kg mass
suspended from it. What is the extension of the spring?
F = |kx| and F = mg
= 0.051 m
Note: A compressed spring has a “-“displacement.
POTENTIAL ENERGY IN A SPRING
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Elastic potential energy is stored in an object if there is no net
deformation.
This means that the object can return to its original form.
Work put into extend a spring, for example, is equal to the work
released by the spring when it returns to its natural shape.
(Some or little heat results.)
Other examples are: trampolines, elastic bands, etc.
A force vs displacement graph for an elastic scenario:
X
Spring breaks
F (N)
Elastic Limit
x (m)
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F = kx in the elastic region only.
The slope is the spring constant k that varies with material.
(N/m)
The energy stored in a spring is equal to the work done to
displace the spring which is represented by the area under the
graph for the elastic region.
Area = (1/2) base height = (1/2) F x = (1/2) kxx = (1/2) kx2
Eel = (1/2) kx2 (again measured in Joules)
Ex 1) Find the elastic potential energy of a spring (k = 160 N/m)
compressed 8.0 cm.
Eel= (1/2) kx2
= (1/2) (160 N/m) (-0.080 m)2
= 0.51 J
Ex 2) A 0.50 kg hockey puck slides at 15.0 m/s. It hits a spring
loaded bumper (k = 360 N/m); how much does the bumper spring
compress at maximum compression?
The puck loses Ek to do work to compress the spring;
- ∆Ek = ∆Eel
- (1/2) mv2 – 0 = 0 - (1/2) kx2
x = ± 0.56 m
As compressed, x = -0.56 m