Download PES 1110 Fall 2013, Spendier Lecture 19/Page 1 Today:

PES 1110 Fall 2013, Spendier
Lecture 19/Page 1
Today:
- Hooke’s Law (chapter section 7.7)
- Elastic potential Energy (chapter section 8.4)
Demo: The effect of the weight hanged on a coil spring on the extension of the spring
A coil spring is hanged with a weight hanger and pointer attached to its end. Different
known weights with equal increments (in this case 500g, 1000g, 1500g, 2000g –
according to the coil's flexibility) are hanged and the readings of the pointer with the
respective weights are marked on a sheet of paper.
The results show that equal weight increments stretch the spring equally (linear
relationship) what we call Hooke's law.
Take in account that this linear relationship between weight and length displacement of
the spring is limited to the working range of the spring (too small weights will not stretch
the spring at all whereas too big weights will compromise its flexibility).
Hooke’s Law:
The force needed to stretch or compress a spring increases linearly with stretching
distance
Hooke’s law uses the stretching distance of the spring. The force increases with
stretching distance. So I will be using a slightly different notation as your textbook. Your
book calls this distance x, but I will call it s for how far the spring has been stretched.
Fsp = k s
where
s = l – l0
k = spring constant, Unit: N/m
(lower case k)
s = stretching distance
(how far has I been stretched)
l0 = equilibrium length
(spring length when spring is unstretched)
l = length of the spring after force is applied
In spring problems put always zero at the position where the spring is unstretched.
PES 1110 Fall 2013, Spendier
Lecture 19/Page 2
Example 1:
A horizontal 50N force is applied to a 100N/m spring whose unstretched length is 0.5m.
What is the spring’s length after the force has been applied?
Example 2:
A 5-kg mass is attached, as shown, to a 100N/m spring whose unstretched length is 0.5m.
If the mass is pushed 0.3 m to the left from the equilibrium position, what is the
magnitude and direction of the force exerted by the spring on the mass?
When you stretch a spring, the spring pulls back. When you compress a spring it will
push back. So springs can push and pull.
PES 1110 Fall 2013, Spendier
Lecture 19/Page 3
Work to stretch a spring:
The spring force is a variable force so how do we find the work needed to stretch the
spring? We need to find the area underneath the force versus position graph.
We have an initial stretching distance s1 and then we will stretch the spring to a distance
s2. The force applied to pull this spring between s1 and s2 increased linearly. Hence we
plot the equation of a straight line thought the origin
Fsp = k s
To find the work, we find the area.
We can subtract the big triangle from the little triangle
1
1
1
1
s2 F2  s1F1  s2 ks2  s1ks1
2
2
2
2
1
1
W  ks22  ks12
2
2
W
(used Hooke’s law: F = k s)
The force changes when you compress or stretch a spring. What happens to work?
Mathematically the difference is that s is positive when you stretch and s is negative
when you compress a spring. What happens when you square this negative number? It
becomes positive. So in terms of work stretch and compress is always the same.
PES 1110 Fall 2013, Spendier
Lecture 19/Page 4
Elastic Potential Energy: Potential energy due to a spring
In spring problems put always zero at the position where the spring is unstretched, since
they have zero potential when they are unstretched.
We pull a box from 0 to s1 to s2. There is potential energy here. When I pull the mass and
I hold it in place, nothing happens until I let go. The mass will move back and will gain
kinetic energy. When I was holding the mass at the rest the kinetic energy was in the
spring stored in from of elastic potential energy. You can think of conservative forces.
While I am pulling, the spring does work. This work is saved. It does not matter if I wait
5 min or 5 years, the instant I let go then you see this work manifest itself in form of
kinetic energy.
Just a while ago, we calculated the force needed to stretch the spring. This corresponds to
the work I have to do when I pull the spring. But I do not do conservative work, the
spring does conservative work. It is the work the spring does on the mass which is saved.
The elastic force exerted by the spring is saved or stored - this force points in the
negative direction!
W
1
1
1
1
s2 F2   s1 F1   s2 ks2  s1 ks1 
2
2
2
2
Elastic work converted to potential energy
1
1 
Wel   ks22  ks12 
 2
2 
But this is exactly what we need since the work done by a conservative force is:
Wel = - ∆ Uel
U el 
1 2
ks elastic potential energy
2
PES 1110 Fall 2013, Spendier
Lecture 19/Page 5
Conservation of Elastic Energy:
If the spring is the only force which is doing work, then the total energy will not change,
it will be conserved.
Ei = Ef
E  K  U el 
1 2 1 2
mv  ks (kinetic energy and elastic potential energy)
2
2
Then the conservation of energy gives:
Ei = Ef
1
1
1
1
mvi 2  ksi 2  mv f 2  ks f
2
2
2
2
(here the mass will not cancel from every term)
Example 3:
A 10 kg mass slides across a frictionless, horizontal floor going 5m/s when it collides
with a k = 500N/m spring. What is the maximum compression of the spring?
Here the mass is slowing down. What is the reason, in form of energy, for the mass to
slow down? There is no friction! When it hits the spring its kinetic energy is changed into
what from? It is changed into elastic potential energy.
There are of course three forces acting on the mass, the normal force, the weight, and the
spring (elastic) force. However, I can apply the conservation of energy here since the
normal force and the weight do no work in this problem. Both are perpendicular to the
motion. To apply the conservation of energy theorem, the spring needs to be the only
force which does work.
Elastic potential energy bungee jumping:
http://stokes.byu.edu/teaching_resources/bungee.html