University Physics (OpenStax) Vol 1, 2021
P2311 Homework 8 (Ch. 8)
Assigned:
Problems
Ch. 8 Concept Q's: 1, 5, 7, 9, 17
Problems: 21, 23, 25, 26, 27, 29, 32, 33, 36, 40, 43
Ch. 8 Concept Qs:
1. The potential energy of a system can be negative because its value is relative to a defined point.
5. The vertical height from the ground to the object
7. A force that takes energy away from the system that can’t be recovered if
we were to reverse the action.
9. The change in kinetic energy is the net work. Since conservative forces are
path independent, when you are back to the same point the kinetic and
potential energies are exactly the same as the beginning. During the trip the
total energy is conserved, but both the potential and kinetic energy change.
17. Four times the original height would double the impact speed.
Ch 8 Problems
21. a. −200J b.−200J c.−100J d.−300J c
23. a. .068 J, b. -.068 J, c. .068 J, d. 068 J, e. -.068 J, f. 46 cm c
25. -120 J c
26. F(x) = - d U/ dx = -a/x^2 + 2b/x^3
27. a) (2a/b)^1/6
b) 0
c) ~ x^6 NO F(x) = 12a/x^13 - 6b/x^7 KEY WRONG
29. 14 m/s or v_f = 13.7 m/s
32. W_res = -190 J
33. Solution manual just says "proof" (not helpful).
Instead use Delta K = - Delta U (cons of mech energy)
1/2 m (v_f^2 - v_i^2) = -mg(h_f - h_i)
v_f^2 - v_i^2 = 20 g
v_f = sqrt( 20*9.8 + 15^2 ) = sqrt(617)=24.8
To show independence from direction try using kinematic equations
for two extreme cases: A) throw straight up, B) throw straight down.
A) v_yf^2 - v_yi^2 = 2 a_y (y_f - y_i)
v_yf^2 - 15^2 = 2 (-9.8) (0 - 20 )
v_yf^2 = 617, so v_yf = 24.8 m/s
B) same as above except v_yi^2 = (-15)^2 instead of (15)^2.
36. a) 4.13 m
b) No, unless there is air resistance and/or the vine has mass.
If these factors exist, then Tarzan won't swing as high on
a longer rope.
40. 6.8 N down ramp (-6.8N i)
43. 3.5 cm check
Check
#21 & #40. -1/2 pt for any conceptual question missing.