Transcript Springs - jpsaos

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.
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)
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
Spring Constant Example
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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?
x1
m1=0.15 kg x1=4.6 cm m2=0.50 kg
x=stretched distance

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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
F1=m1g
F=(m1+m2)g
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)
W= ½
x
kx2
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
PEs= ½ kx2
Elastic potential energy gained
by a spring.
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
x1
W= ½ kx22
From position 2 to 3:
W23 = W3-W2= ½ kx32 - ½ kx22= ½ k(x32-x22)
x2
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!
x3
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
Conservation of Mechanical Energy
(Springs)
v1
m
Time 1
x1
v2
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
m
Time 2
x2
Springs Conservation of Mechanical
Energy Equations
 ME1=ME2
 PE1+KE1=PE2+KE2
½
2
kx1 +
½
2
mv1 =
½
2
kx2 +
½
2
mv2
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