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Hooke’s
Law
• Hooke's Law gives the force acting on a
spring or other elastic material when it has
been stretched or compressed
• The direction of the force is opposite to the
direction of stretch
Weight of object stretches spring
Restoring force acts up
force
k spring constant (N/m)
of spring
(change in length)
s
s
F k x
F restoring
x displacement
Hooke’s Law (AP)
• Fs = restoring force
• x = displacement
sF k x
• F is proportional to x
sF k x
Is the equation of a line
Example
• A 400 g mass is hung form the lower end
of a spring. The spring stretches 0.200 m.
Calculate k.
Solution
m/N.k
m.
kg/N.kg.k
x
gm
x
Fk
xkF
s
s
619
2000
8194000
Example
• A 150 g mass is attached to one end of a
horizontal spring (k = 44.3 N/m) and the
spring is stretched 0.104 m.
a) Determine the maximum acceleration
when the mass is released.
x
Solution
2
44.3 / 0.104
0.150
30.7 / left
s netF F
k x m a
k xa
m
N m ma
kg
a m s
Apply Newton’s
2nd Law
b) Determine the acceleration of the object
when it is 0.055 m from the equilibrium
position
2
44.3 / 0.055
0.150
16.2 / left
s netF F
k x m a
k xa
m
N m ma
kg
a m s
• The magnitude of the acceleration is
constantly changing since the restoring
force changes as x changes
• Non-uniform accelerated motion means
you can’t use the kinematics equations
from unit 1
Example
• A spring is connected to a box and its k
determined. The spring is then cut in half
and connected to the box as shown.
Describe the value of k for the two springs
Solution
• It's 2k, since each spring now only extends
half the displacement of the original spring
when subjected to the same force.
More than One Spring
• Two springs are
connected vertically in
series and a mass is
suspended from the
lower end. Derive an
expression for the
effective spring
constant.
More than One Spring
• F1 = -k1x1
• F2 = -k2x2
• The forces on each
spring are equal. (ignore
the spring masses)
• -k1x1 = -k2x2
More than One Spring
• -k1x1 = -k2x2
• F = -keff(x1 + x2)
• -keff(x1 + x2) = -k1x1 + -k2x2
2 21
1
k xx
k
More than One Spring
𝑘𝑒𝑓𝑓𝑘2𝑥2𝑘1
+ 𝑥2 = 𝑘2𝑥2
Divide by x2
• 𝑘𝑒𝑓𝑓𝑘2
𝑘1+ 1 = 𝑘2
More than One Spring
• 𝑘𝑒𝑓𝑓𝑘2
𝑘1+ 1 = 𝑘2
• 𝑘𝑒𝑓𝑓𝑘2+𝑘1
𝑘1= 𝑘2
• 𝑘𝑒𝑓𝑓 =𝑘2𝑘1
𝑘2+𝑘1
1 2
1 1 1
effk k k
More than One Spring
• Two springs are
connected as shown and
a force is applied. What
is the spring constant of
the system? (connected
in parallel)
More than One Spring
• The value of x is the
same for both springs
• F = -k1x + -k2x
• F = -(k1 + k2)x
• keff = k1 + k2
Example • Two identical unstrained springs (k = 15
N/m) are connected to an 1.0 kg object as
shown. If the object is moved 4.0 cm to the
left and released, what is the acceleration
magnitude of the system?
Spring Potential Energy
• the energy stored
in a spring
depends on the
displacement
• the work done to stretch the spring = area
under graph
2
2
1
2
1
2
1
2
1
kxE
xkxE
FxE
abEwork
AP
• Us = ½ kx2
Practice
• P 301: 1, 2, 3, 4, 5