fisica1-youn-e-freedman-exercicios-resolvidos

7.85: a) For the given proposed potential U ( x),
−
dU = −kx
+ F , so this is a possible
2
potential function. For this potential, U (0) = − F 2k
, not zero. Setting the zero of
potential is equivalent to adding a constant to the potential; any additive constant will not
change the derivative, and will correspond to the same force. b) At equilibrium, the force
is zero; solving − kx + F = 0 for x gives x
0
= F k . U ( x ) 2
0
= − F k , and this is a
minimum of U, and hence a stable point.
c)
dx
d) No; F 0 at only one point, and this is a stable point. e) The extreme values of x
tot
=
correspond to zero velocity, hence zero kinetic energy, so U ( x ) ±
= E , where x
±
are the
extreme points of the motion. Rather than solve a quadratic, note that
1 2 2
2
k( x − F k)
− F k , so U ( x ) ±
= E becomes 2
2
1 ⎛ F ⎞ F
k⎜
x±
− ⎟ − F k =
2 ⎝ k ⎠ k
F F
x±
− = ± 2 ,
k k
F F
x+
= 3 x−
= − .
k k
f) The maximum kinetic energy occurs when U (x)
is a minimum, the point x = F k
found in part (b). At this point
v = 2F mk .
2
2
2
K = E −U
= ( F k)
− ( − F k)
= 2F
k , so
0