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The Spring
Consider a spring, which we apply a force FA to
either stretch it or compress it
FA
x
unstretched
-FA
-x
x=0
FA  kx
k is the spring
constant, units of
N/m, different for
different materials,
number of coils
From Newton’s 3rd Law, the spring exerts a force
that is equal in magnitude, but opposite in
direction
Hooke’s Law for the restoring
force of an ideal spring. It is
s
a conservative force.
F  kx
Work and Potential Energy of a Spring
 We know that work equals (force times
displacement)
s  x f  xi
But the force is not constant
xi
xf
Fs ,i  kxi ,
Fs , f  kx f
Take the average force
Fs 
x=0
Fs ,i  Fs , f
2
Fs   12 k ( x f  xi )
Then the work done by the spring is
Ws  Fs cos  s  Fs cos 0 s
2
2
1
1
  2 k ( x f  xi )( x f  xi )   2 k ( x f  xi )
2
2
1
1
Ws  2 kxi  2 kx f  U s ,i  U s , f

U elastic  U s  kx
1
2
2
Units of N/m m2
=Nm=J
 Total potential energy is
Utotal  U g  U s  mgh  kx
1
2
2
Example Problem
A block (m = 1.7 kg) and a spring (k = 310 N/m)
are on a frictionless incline ( = 30°). The spring is
compressed by xi = 0.31 m relative to its
unstretched position at x = 0 and then released.
What is the speed of the block when the spring is
still compressed by xf = 0.14 m?
x=0
xi
x=0 xf


Given: m=1.7 kg, k=310 N/m, =30°, xi=0.31 m,
xf=0.14 m, frictionless
Method: no friction, so we can use conservation
of energy
Initially
E  mv  mgh  kx
vi  0, hi  xi sin 
1
2
2
1
2
Ei  mgxi sin   kx
1
2
2
i
2
Finally
h f  x f sin  , find v f
2
2
1
1
E f  2 mv f  mgx f sin   2 kx f
E f  Ei
2
2
1
1
2 mv f  mgx f sin   2 kx f
2
1
 mgxi sin   2 kxi
2
2
2
1
1
2 mv f  mg ( xi  x f ) sin   2 k ( xi  x f )
k 2
2
v f  2 g ( xi  x f ) sin   ( xi  x f )
m
310
v f  2(9.8)(0.31  0.14) sin 30 
(0.312  0.14 2 )
1.7
v f  1.666  13.95  3.95 ms

Interesting to plot the potential energies
Energy
K

Ug
xf
E
Utotal
Us
xi
xf
xi
xf
xi