1. Springs
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.

2. Force Applied to a Spring
x0 (unstretched position) F x1
F=k(x1-x0)
x0 is at the original unstretched position of the spring endpoint, x1= is the position of the spring after being stretched. Commonly written as F=kx, x=x1-x0
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)

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

4. Spring Constant Example
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?
m1=0.15 kg x1=4.6 cm m2=0.50 kg x=stretched distance
k=F1/x1=(.15kg)(9.8m/s2)/.046m=32 N/m
x=F/k=(m1+m2)g/k
=(0.15 kg+0.50 kg)(9.8 m/s2)/(32 N/m)=
=0.20 m=20 cm x1
F1=m1g
F=(m1+m2)g

5. Work Accomplished on a Spring (Energy stored in a Spring)
Work is the area under a Force-Displacement graph. F
W= ½ xF= ½ x(kx)
The work accomplished in moving a spring from a zero reference to a position x.
W= ½ kx2 x
Since work is a transfer of energy, then potential energy
is gained by a spring elongated from its reference position
PEs= ½ kx2
Elastic potential energy gained
by a spring.

6. Work Moving a Spring between Two Locations
The work (stored energy) moving the spring
From position 1 to 2:
W 12= W2-W1= ½ kx22- ½ kx12= ½ k(x22-x12)
x1=0 for an unstretched spring
W= ½ kx22
From position 2 to 3:
W23 = W3-W2= ½ kx32 - ½ kx22= ½ k(x32-x22)
The work is moving the spring from
position 1 to 2 or position 2 to 3 is NOT:
W12= ½ k(x2-x1)2
W23= ½ k(x3-x2)2
Common mistake/misconception! x1 x2 x3

7. 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 =W1-W0 = ½ kx12- ½ kx02= ½ k(x12-x02)=
½ k(x12-0)= ½ kx12= 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 =W2-W1= ½ kx22- ½ kx12= ½ k(x22-x12)=
½ (750N/m)[(.05m)2-(.02m)2] =0.79 J

8. Conservation of Mechanical Energy (Springs)
Time 1 x1
x1=spring endpoint position at time 1
v1=mass velocity at time 1
x2=spring endpoint position at time 2
v2=mass velocity at time 2 v2 m
Time 2 x2

9. Springs Conservation of Mechanical Energy Equations
ME1=ME2
PE1+KE1=PE2+KE2
½ kx12+ ½ mv12= ½ kx22+ ½ mv22

10. Conservation of Mechanical Energy for Springs Example Problem
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?
ME1=ME2
PE1+KE1=PE2+KE2
½ kx12+ ½ mv12= ½ kx22+ ½ mv22
0+ ½ (10 kg)(1.0 m/s)2= ½ (800 N/m)x22+0
x2= 0.11 m = 11 cm