Reading a Potential Energy Curve
Once again we consider a particle that is part of a system in which a conservative
force acts. This time suppose that the particle is constrained to move along an
x axis while the conservative force does work on it. We want to plot the potential
energy U(x) that is associated with that force and the work that it does, and then
we want to consider how we can relate the plot back to the force and to the kinetic
energy of the particle. However, before we discuss such plots, we need one more
relationship between the force and the potential energy.
Finding the Force Analytically
Equation 8-6 tells us how to find the change ⌬U in potential energy between two
points in a one-dimensional situation if we know the force F(x). Now we want to
187
8-3 READING A POTENTIAL ENERGY CURVE
To show both kinds of mechanical energy, we have
K b ϩ U b ϭ K t ϩ U t ,
( 8 - 2 0 )
or
Dividing by m and rearranging yield
Putting v t ϭ 0 and y t Ϫ y b ϭ h leads to
(Answer)
ϭ 13 m/s.
v b ϭ 12gh ϭ 1(2)(9.8 m/s
2
)(8.5 m)
v b
2 ϭ v t
2 ϩ 2g(y t Ϫ y b ).
1
2 mv b
2 ϩ mgy b ϭ
1
2 mv t
2 ϩ mgy t .
Additional examples, video, and practice available at WileyPLUS
8-3 READING A POTENTIAL ENERGY CURVE
After reading this module, you should be able to . . .
8.07 Given a particle’s potential energy as a function of its
position x, determine the force on the particle.
8.08 Given a graph of potential energy versus x, determine
the force on a particle.
8.09 On a graph of potential energy versus x, superimpose a
line for a particle’s mechanical energy and determine the
particle’s kinetic energy for any given value of x.
8.10 If a particle moves along an x axis, use a potentialenergy graph for that axis and the conservation of mechanical energy to relate the energy values at one position to
those at another position.
8.11 On a potential-energy graph, identify any turning points
and any regions where the particle is not allowed because
of energy requirements.
8.12 Explain neutral equilibrium, stable equilibrium, and
unstable equilibrium.
Learning Objectives
Key Ideas
● If we know the potential energy function U(x) for a system
in which a one-dimensional force F(x) acts on a particle, we
can find the force as
● If U(x) is given on a graph, then at any value of x, the force
F(x) is the negative of the slope of the curve there and the
F(x) ϭ Ϫ
dU(x)
dx
.
kinetic energy of the particle is given by
K(x) ϭ E mec Ϫ U(x),
where E mec is the mechanical energy of the system.
● A turning point is a point x at which the particle reverses its
motion (there, K ϭ 0).
● The particle is in equilibrium at points where the slope of
the U(x) curve is zero (there, F(x) ϭ 0).
This is the same speed that the child would reach if she fell
8.5 m vertically. On an actual slide, some frictional forces
would act and the child would not be moving quite so fast.
Comments: Although this problem is hard to solve directly
with Newton’s laws, using conservation of mechanical energy makes the solution much easier. However, if we were
asked to find the time taken for the child to reach the bottom of the slide, energy methods would be of no use; we
would need to know the shape of the slide, and we would
have a difficult problem.
Once again we consider a particle that is part of a system in which a conservative
force acts. This time suppose that the particle is constrained to move along an
x axis while the conservative force does work on it. We want to plot the potential
energy U(x) that is associated with that force and the work that it does, and then
we want to consider how we can relate the plot back to the force and to the kinetic
energy of the particle. However, before we discuss such plots, we need one more
relationship between the force and the potential energy.
Finding the Force Analytically
Equation 8-6 tells us how to find the change ⌬U in potential energy between two
points in a one-dimensional situation if we know the force F(x). Now we want to
187
8-3 READING A POTENTIAL ENERGY CURVE
To show both kinds of mechanical energy, we have
K b ϩ U b ϭ K t ϩ U t ,
( 8 - 2 0 )
or
Dividing by m and rearranging yield
Putting v t ϭ 0 and y t Ϫ y b ϭ h leads to
(Answer)
ϭ 13 m/s.
v b ϭ 12gh ϭ 1(2)(9.8 m/s
2
)(8.5 m)
v b
2 ϭ v t
2 ϩ 2g(y t Ϫ y b ).
1
2 mv b
2 ϩ mgy b ϭ
1
2 mv t
2 ϩ mgy t .
Additional examples, video, and practice available at WileyPLUS
8-3 READING A POTENTIAL ENERGY CURVE
After reading this module, you should be able to . . .
8.07 Given a particle’s potential energy as a function of its
position x, determine the force on the particle.
8.08 Given a graph of potential energy versus x, determine
the force on a particle.
8.09 On a graph of potential energy versus x, superimpose a
line for a particle’s mechanical energy and determine the
particle’s kinetic energy for any given value of x.
8.10 If a particle moves along an x axis, use a potentialenergy graph for that axis and the conservation of mechanical energy to relate the energy values at one position to
those at another position.
8.11 On a potential-energy graph, identify any turning points
and any regions where the particle is not allowed because
of energy requirements.
8.12 Explain neutral equilibrium, stable equilibrium, and
unstable equilibrium.
Learning Objectives
Key Ideas
● If we know the potential energy function U(x) for a system
in which a one-dimensional force F(x) acts on a particle, we
can find the force as
● If U(x) is given on a graph, then at any value of x, the force
F(x) is the negative of the slope of the curve there and the
F(x) ϭ Ϫ
dU(x)
dx
.
kinetic energy of the particle is given by
K(x) ϭ E mec Ϫ U(x),
where E mec is the mechanical energy of the system.
● A turning point is a point x at which the particle reverses its
motion (there, K ϭ 0).
● The particle is in equilibrium at points where the slope of
the U(x) curve is zero (there, F(x) ϭ 0).
This is the same speed that the child would reach if she fell
8.5 m vertically. On an actual slide, some frictional forces
would act and the child would not be moving quite so fast.
Comments: Although this problem is hard to solve directly
with Newton’s laws, using conservation of mechanical energy makes the solution much easier. However, if we were
asked to find the time taken for the child to reach the bottom of the slide, energy methods would be of no use; we
would need to know the shape of the slide, and we would
have a difficult problem.