fisica1-youn-e-freedman-exercicios-resolvidos

∂ U a b
7.37: a) F − = 12
13
− 6
7
.
r
=
∂r
r
r
6
b) Setting F
r
= 0 and solving for r gives r (2 )
1/
min
= a b . This is the minimum of
potential energy, so the equilibrium is stable.
c)
a b
U ( rmin)
= −
12 6
r r
min
a
=
((2a
/ b)
b
−
((2a
/ b)
2 2 2
ab b b
= − = − .
2
4a
2a
4a
To separate the particles means to remove them to zero potential energy, and requires the
2
negative of this, or E
0
= b 4a.
d) The expressions for E
0
and r min
in terms of a and b
are
2
b 6 2a
E
0
= rmin
= .
4a
b
6
Multiplying the first by the second and solving for b gives b = 2E
r , and substituting
this into the first and solving for a gives
a = (1.54×
10
−18
a =
J)(1.13×
10
min
1/ 6
)
E r 12
0 min
12
1/ 6
)
6
0 min
. Using the given numbers,
= 6.68×
10
J ⋅ m
−18
−10
6
−78
6
b = 2(1.54×
10 J)(1.13×
10 m) = 6.41×
10 J ⋅ m .
(Note: the numerical value for a might not be within the range of standard calculators,
and the powers of ten may have to be handled seperately.)
−10
m)
12
−138
12
dU
7.38: a) Considering only forces in the x-direction, Fx
= −
dx
, and so the force is zero
when the slope of the U vs x graph is zero, at points b and d. b) Point b is at a potential
minimum; to move it away from b would require an input of energy, so this point is
stable. c) Moving away from point d involves a decrease of potential energy, hence an
increase in kinetic energy, and the marble tends to move further away, and so d is an
unstable point.