K
11
=
E
max
–
U
11
= (14.0 – 12.0) J = 2.00 J.
(g) Now we have
W
=
F
4
Δ
x =
(–1.00 N)(1.00 m) = –1.00 J, so the potential energy at
12.0 m
x
=
is
U
12
= 1.00 J +
U
11
= (1.00 + 12.0) J = 13.0 J.
(h) Thus, the kinetic energy at
x
= 12.0 m is
K
12
=
E
max
–
U
12
= (14.0 – 13.0) = 1.00 J.
(i) There is no work done in this interval (from
x
= 12.0 m to
x
= 13.0 m) so the answers
are the same as in part (g):
U
12
= 13.0 J.
(j) There is no work done in this interval (from
x
= 12.0 m to
x
= 13.0 m) so the answers
are the same as in part (h):
K
12
= 1.00 J.
(k) Although the plot is not shown here, it would look like a “potential well” with
piecewisesloping sides: from
x
= 0 to
x
= 2 (SI units understood) the graph if
U
is a
decreasing line segment from 11 to 5, and from
x
= 2 to
x
= 3, it then heads down to zero,
where it stays until
x
= 8, where it starts increasing to a value of 12 (at
x
= 11), and then
in another positiveslope line segment it increases to a value of 13 (at
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics, Energy, Force, Potential Energy, Work

Click to edit the document details