SME
University Physics I: Mechanics · Problem type
Force and equilibrium from U(x) = ax³ − bx²
Use this for “A particle has potential energy U(x) = 2x³ − 6x² (U in J, x in m). Find the force on it at x = 0.5 m, and where it can rest in stable equilibrium.”
Chart
Local maximum at (0, 0)
Local minimum at (2, −4)
U(1) = −2
Tangent at x = 1: slope U′(1) = −3, y = −3x + 1
This is an example. Type your own values to start a new problem.
U(x)
=
a
x³
−
b
x²
F(
x (m)
)
=
F (N)
x
stable
=
xₛ (m)
U
Potential energy
worked out (not typed) · J
Uₛ
Energy at the stable position
worked out (not typed) · J
Formulas
U
=
a
×
x
3
−
b
×
x
2
−2
=
1
×
1
3
−
3
×
1
2
F
=
−3
×
a
×
x
2
+
2
×
b
×
x
3
=
−3
×
1
×
1
2
+
2
×
3
×
1
xₛ
=
2
×
b
3 ×
a
2
=
2
×
3
3 × 1
Uₛ
=
a
×
xₛ
3
−
b
×
xₛ
2
−4
=
1
×
2
3
−
3
×
2
2
Assumptions
•
The force is conservative, with potential energy U(x) = ax³ − bx²: a in J/m³ and b in J/m², both positive.
•
The force is the downhill slope of U: F = −dU/dx, so F > 0 pushes toward +x.
•
Equilibrium is where F = 0. It is stable at a minimum of U (U″ > 0) and unstable at a maximum.
•
U″ = 6ax − 2b, so the minimum at xₛ = 2b ÷ (3a) is stable (U″ = 2b) and x = 0 is unstable (U″ = −2b).
Step-by-step
WE KNOW
a = 1 J/m³, b = 3 J/m², x = 1 m
FIND
potential energy (U), force (F), stable position (xₛ), energy at the stable position (Uₛ)
Step 1 · Find potential energy (U)
Use U = a × x³ − b × x²
Write
U
(
x
)
with
its
numbers,
then
put
the
position
in,
one
term
at
a
time.
U
=
x
3
−
3
x
²
U
=
1
×
1
3
−
3
×
1
2
U
=
1
−
3
U
=
−2
J
Step 2 · Find force (F)
Use F = −3 × a × x² + 2 × b × x
F
=
−dU/dx.
By
the
power
rule
dU/dx
=
3ax²
−
2bx;
change
its
sign
and
put
x
in.
F
=
−
3
x
²
+
6
x
F
=
−3
×
1
×
1
2
+
2
×
3
×
1
F
=
−3
+
6
F
=
3
N
Step 3 · Find stable position (xₛ)
Use xₛ = 2 × b ÷ (3 × a)
F
=
0
where
3ax²
−
2bx
=
x
(3ax
−
2
b
)
=
0:
at
x
=
0
and
at
x
=
2
b
3
a
.
U
″
=
2
b
>
0
at
the
second,
a
minimum.
xₛ
=
2
b
3
a
U
(
x
)
=
x
3
−
3
x
²
U
′(
x
)
=
3
x
²
−
6
x
=
3
x
(
x
−
2)
xₛ
=
2
×
3
3 × 1
xₛ
=
6
3
U
″(
x
)
=
6
x
−
6
U
″(2)
=
6
×
2
−
6
=
6,
above
0:
a
minimum
xₛ = 2 m
Step 4 · Find energy at the stable position (Uₛ)
Use Uₛ = a × xₛ³ − b × xₛ²
Put
the
stable
position
into
U
(
x
):
the
bottom
of
the
well.
Uₛ
=
1
×
2
3
−
3
×
2
2
Uₛ = −4 J
Check
Put the numbers back in. Both sides should match.
−2
=
1
×
1
3
−
3
×
1
2
✓
3
=
−3
×
1
×
1
2
+
2
×
3
×
1
✓
2
=
2
×
3
3 × 1
✓
−4
=
1
×
2
3
−
3
×
2
2
✓
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Energy down a rough slope: heat and the end speed
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