A coiled mechanical device that stores elastic potential energy by compression or elongation
Elastic Potential Energy – The energy stored in an object due to a deformation.

x
0
(unstretched position)
F x
1
F=k(x
1
•x
0
-x
0
) is at the original unstretched position of the spring endpoint, x
1
= is the position of the spring after being stretched. Commonly written as
F=kx , x=x
1
-x
0
• x is the amount of compression or elongation from the equilibrium position measured in meters.
•k – spring constant (force constant) – a measure of the resistance of the spring to deformation.
•k is measured in N/m
•The spring scales used in class have a spring constant of 350 N/m (a moderate spring constant)

• The resistive force applied by the spring to return to its original shape is called the restoring force (F s
).
• F s
=-kx
. The restoring force is equal in magnitude and opposite in direction to the applied force.
• F=-F s
•The restoring force applied by the spring is called Hooke’s Law
.
Hooke’s Law:
F s
= -kx
F
S
F







A 0.15 kg mass is attached to a vertical spring and hangs at rest a distance of 4.6 cm below its original position. An additional 0.50 kg mass is then suspended from the first mass and allowed to descend to a new equilibrium position. What is the total extension of the spring?
m
1
=0.15 kg x
1
=4.6 cm x=stretched distance m
2
=0.50 kg k=F
1
/x
1
=(.15kg)(9.8m/s 2 )/.046m=32 N/m x=F/k=(m
1
+m
2
)g/k
=(0.15 kg+0.50 kg)(9.8 m/s 2 )/(32 N/m)=
=0.20 m=20 cm x
1
F
1= m
1 g
F=(m
1
+m
2
)g

F

Work is the area under a Force-Displacement graph.
x
W= ½ xF= ½ x(kx)
W= ½ kx 2
The work accomplished in moving a spring from a zero reference to a position x.
Since work is a transfer of energy, then potential energy is gained by a spring elongated from its reference position
PE s
= ½ kx 2
Elastic potential energy gained by a spring.

Work Moving a Spring between Two Locations x
1 x
2
The work (stored energy) moving the spring
From position 1 to 2:
W
1  2
= W
2
-W
1
= ½ kx
2
2 ½ kx
1
2 = ½ k(x
2
2 -x
1
2 ) x
1
=0 for an unstretched spring
W= ½ kx
2
2
From position 2 to 3:
W
2  3
= W
3
-W
2
= ½ kx
3
2 ½ kx
2
2 = ½ k(x
3
2 -x
2
2 ) x
3
The work is moving the spring from position 1 to 2 or position 2 to 3 is NOT :
W
1  2
= ½ k(x
2
-x
1
) 2
W
2  3
= ½ k(x
3
-x
2
) 2
Common mistake/misconception!

Example of Work in Moving a Spring between Two Locations
How much work is required to move a spring with a spring constant of 750 N/m from its unstretched position to 2.0 cm?
W= W
0  1
½ k(x
1
2
=W
1
-W
0
-0)= ½ kx
1
2
= ½ kx
1
2 - ½ kx
0
2 = ½ k(x
1
2 -x
0
2 )=
= ½ (750 N/m)(.02m) 2 =0.15 J
How much work is required to move the same spring an additional 3.0cm?
W=W
1  2
=W
2
-W
1
= ½ kx
2
2 - ½ kx
1
2 = ½ k(x
2
2 -x
1
2 )=
½ (750N/m)[(.05m) 2 -(.02m) 2 ]
=0.79 J

v
1 m
Time 1 x
1 m v
2 x
1 v
1 x v
2
2
=spring endpoint position at time 1
=mass velocity at time 1
=spring endpoint position at time 2
=mass velocity at time 2
Time 2 x
2


ME
1
=ME
2

PE
1
+KE
1
=PE
2
+KE
2

½ kx
1
2 + ½ mv
1
2 = ½ kx
2
2 + ½ mv
2
2

Conservation of Mechanical Energy for Springs

A 10 kg mass traveling at 1.0 m/s on a frictionless surface compresses an originally undeformed spring with a constant of 800 N/m until it stops. What is the distance that the spring is compressed?


ME
1
=ME
2
PE
1
+KE
1
=PE
2
+KE
2

½ kx
1
2 + ½ mv
1
2 = ½ kx
2
2 + ½ mv
2
2
0+ ½ (10 kg)(1.0 m/s) 2 = ½ (800 N/m)x
2
2 +0 x
2
= 0.11 m = 11 cm