Problem#1
A right circular cone can be balanced on a horizontal surface in three different ways. Sketch these three equilibrium configurations, and identify them as positions of stable, unstable, or neutral equilibrium.
Answer:
Problem#2
For the potential energy curve shown in Figure 2, (a) determine whether the force Fx is positive, negative, or zero at the five points indicated. (b) Indicate points of stable, unstable, and neutral equilibrium. (c) Sketch the curve for Fx versus x from x = 0 to x = 9.5 m.
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| Fig.2 |
Answer:
(a) Fx is zero at points A, C and E; Fx is positive at point B and negative at point D.
(b) A and E are unstable, and C is stable.
(c)
Problem#3
A particle moves along a line where the potential energy of its system depends on its position r as graphed in Figure 4. In the limit as r increases without bound, U(r) approaches +1 J. (a) Identify each equilibrium position for this particle. Indicate whether each is a point of stable, unstable, or neutral equilibrium. (b) The particle will be bound if the total energy of the system is in what range?
Now suppose that the system has energy –3 J. Determine (c) the range of positions where the particle can be found, (d) its maximum kinetic energy, (e) the location where it has maximum kinetic energy, and (f) the binding energy of the system—that is, the additional energy that it would have
to be given in order for the particle to move out to r → ∞.
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| Fig.4 |
Answer:
(a) There is an equilibrium point wherever the graph of potential energy is horizontal:
- At r = 1 5 . mm and 3.2 mm, the equilibrium is stable.
- At r = 2 3 . mm, the equilibrium is unstable.
- A particle moving out toward r → ∞ approaches neutral equilibrium.
(b) The system energy E cannot be less than –5.6 J. The particle is bound if –5.6 J ≤ E < 1 J
(c) If the system energy is –3 J, its potential energy must be less than or equal to –3 J. Thus, the
particle’s position is limited to 0.6 mm ≤ r ≤ 3.6 mm.
(d) K + U = E. Thus Kmax = E – Umin = –3 J – (–5.6 J) = 2.60 J
(e) Kinetic energy is a maximum when the potential energy is a minimum, at r = 1.5 mm.
(f) –3 + W = 1 J. Hence, the binding energy is W = 4 J
Problem#5
A particle of mass 1.18 kg is attached between two identical springs on a horizontal frictionless tabletop. The springs have force constant k and each is initially unstressed. (a) If the particle is pulled a distance x along a direction perpendicular to the initial configuration of the springs, as in Figure 5, show that the potential energy of the system is
U(x) = kx2 + 2kL[L – (x2 + L2)1/2]
(Hint: See Problem 58 in Chapter 7.) (b) Make a plot of U(x) versus x and identify all equilibrium points. Assume that L = 1.20 m and k = 40.0 N/m. (c) If the particle is pulled 0.500 m to the right and then released, what is its speed when it reaches the equilibrium point x = 0?
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| Fig.5 |
Answer:
(a) The new length of each spring is (x2 + L2)1/2, so its extension is (x2 + L2)1/2 – L and the force it exerts is
k{(x2 + L2)1/2 – L }
toward its fixed end. The y components of the two spring forces add to zero. Their x components add to
Fx = –2k{(x2 + L2)1/2 – L}[x/(x2 +
Fx = –2kx{1 – L/(x2 + L2)1/2}
Choose U = 0 at x = 0 . Then at any point the potential energy of the system is
U = –∫Fxdx
= ∫0x[2kx{1 – L/(x2 + L2)1/2}]dx
= ∫0x(2kx)dx – ∫0x{2xkL/(x2 + L2)1/2}dx
= kx2│0x – 2kL(x2 + L2)1/2│0x
U = kx2 + 2kL{L – (x2 + L2)1/2}
(b) Given: L = 1.20 m, k = 40.0 N/m, then
U(x) = 40.0x2 + 96.0{1.20 – (x2 + 1.44)}
For negative x, U(x) a f has the same value as for positive x. The only equilibrium point (i.e., where Fx = 0) is x = 0.
(c) We us
∆Emech = ∆K + ∆U = Kf – Ki + Uf – Ui
0 = ½ mv2 – 0 + 0 – Ui
0.400 J = ½ (1.18 kg)v2
v = 0.823 m/s
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