Transcript Mechanics 105 chapter 7

Mechanics 105
Potential energy (chapter seven)
Potential energy of a system
 The isolated system
 Conservative and nonconservative forces
 Conservative forces and potential energy
 The nonisolated system in steady state
 Potential energy for gravitational and electrostatic
forces
 Energy diagrams and stability

Mechanics 105
Potential energy of a system
 Objects and force internal to system
 External work done on a system that does not
change the kinetic or internal energy – change in
potential energy
 Gravitational potential energy - the work done by
an external force raising an object from ya to yb is:
 
W  F  r  mg( yb  ya )
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Potential energy
This work is a transfer of energy to the system. We
can define the quantity Ugmgy to be the
gravitational potential energy
The work done on the system then gives the
change in Ug: W= Ug
The change in energy is the key thing – to do
problems, you need to first define a reference
location (height)
Only depends on height – not on horizontal
displacement
Mechanics 105
Isolated systems
Consider the system of an object only in the earth’s
gravitational field, falling from yb to ya.
In free fall, the work done by gravity is mg(yb-ya), which
results in a change in the kinetic energy (work-kinetic
energy theorem) K.
This work equals -Ug, the change in the gravitational
potential energy of the (earth + object) system. The
earth’s kinetic energy will not change, so the change in
kinetic energy of the (earth + object) system is just the
change of the KE of the object.
This gives the result: K= -Ug, or K+Ug=0
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Isolated systems
 We can write this as a continuity equation for the
mechanical energy
Emech=K+Ug
 Emech=0, or Ki+Ui=Kf+Uf
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Example – object in free fall
 Consider earth and object as system
 Object dropped from height y=h (y=0 is
defined as height at which Ug=0)
 At height y, what is the speed?
Mechanics 105
Example – object in free fall
Initial energy = K+Ug=mgh
Final energy (at any point y) =
mgy+½mv2
mgh=mgy+½mv2  v=(2g(h-y))½
y=h
Ug=mgh
K=0
y
Ug=mgy
K=½mv2
y=0
Ug=0
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 Conceptest
 Demo
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Example
If m1>m2, How fast will m1 be going when it hits the
floor?
Start: K+Ug=m1gh
End: ½m1v2+m2gh
m1
m1gh=½m1v2+m2gh
v=[2gh(m1-m2)/m1]½

m2
Mechanics 105
Conservative and nonconservative
forces
Conservative forces
- Forces internal to system that cause no
transformation of mechanical to internal energy
- Work done is path independent
- Work done over closed path = 0
- Examples: gravitational, elastic
Mechanics 105
Conservative and nonconservative
forces
Potential energy of a spring
Us½kx2
Gravitational potential energy
Ugmgh
New formulation of Work-KE thm:
K+U+Eint=constant
Conservation of energy
Mechanics 105
Example
Motion on a curved track
Child slides down an irregular frictionless track (total height h) ,
starting from rest. What is the speed at the bottom?
Ki+Ui= Kf+Uf
0+mgh=½mv2+0  v=(2gh)½
Mechanics 105
Conservative forces and potential
energy
Since the work done by a conservative force can
be written as W=-U
We can express a differential amount of work done
as dW=-dU=F·dr
From this we can see that a conservative force can
be written as Fx=-dU/dx (F=-U)
e.g. Fg=-dUg/dy=-d(mgy)/dy=-mg
Mechanics 105
The nonisolated system in steady state
Conservation of energy holds regardless of whether
the system is isolated or not. For a nonisolated
system, the net energy change can still be zero if
the amount of energy entering equals the amount
leaving the system.
Mechanics 105
Potential energy for gravitational and
electrostatic forces
Gravitational force between two masses (m1, m2) separated
by a distance r

Gm1m2
Fg12  
r̂12
2
r
This gives a general form for the gravitational potential
energy of:
Gm1m2
Ug  
r
And for the electrostatic potential energy
k e q1q2
Ue 
r
Mechanics 105
Example
Mechanics 105
Example
Mechanics 105
Energy diagrams and stability
Since the potential energy associated with a
conservative force can be written Fx=-dU/dx, a plot of
U vs. x can tell us something about how a system
will behave as a function of position.
For relative minima of U vs. x, there will be no force –
we call these points stable equilibria.
For relative maxima of U vs. x, there will also be no
force, but for small displacements away from this
point, the force will be away from the equilibrium
point – we call these points unstable equilibria.
Mechanics 105
Energy diagrams and stability
Example: mass on a spring – stable equilibrium point at x=0
4.5
Us (J)
4
3.5
3
2.5
2
1.5
1
0.5
0
-20
-10
0
position (m)
10
20
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Energy diagrams and stability
Mechanics 105
Mechanics 105
Mechanics 105