ch08-p117 - 117(a The remark in the problem statement that...

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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 piecewise-sloping 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 positive-slope line segment it increases to a value of 13 (at
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ch08-p117 - 117(a The remark in the problem statement that...

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