Question

A particle moves along a line where the potential energy of its system depends on its position r as graphed in Figure P8.46. 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→ ∞ .

Answer 1

a) The equilibrium positions for this particle can be found at the positions where the slope of the potential energy graph (U(r)) is zero.

By looking at the graph, we can find three equilibrium positions at r ~ 0, r ~ 40, and r ~ 70. To determine whether each of these positions is stable, unstable, or neutral, we can examine the curvature of the graph around these positions.

- At r ~ 0, the graph is concave up, indicating that this is a point of stable equilibrium.
- At r ~ 40, the graph is concave down, indicating that this is a point of unstable equilibrium.
- At r ~ 70, the graph is concave up, indicating that this is a point of stable equilibrium.

b) The particle will be bound if the total energy of the system is less than +1 J, which is the potential energy for r → ∞. If the total energy were equal to or greater than +1 J, the particle could gain enough kinetic energy to move out to r → ∞.

c) With a total energy of -3 J, the particle can be found in positions where the potential energy is less than or equal to -3 J. Looking at the graph, this corresponds to the range of positions between r ~ 10 and r ~ 50.

d) The maximum kinetic energy will occur when the particle is at its lowest potential energy. This occurs at the bottom of the potential well, where U(r) ≈ -6 J. Since the total energy is conserved, at this point, the kinetic energy will be:

K_max = E_total - U(r) = -3 J - (-6 J) = 3 J

e) The particle has maximum kinetic energy when it is at the bottom of the potential well, which occurs at a location of r ~ 30.

f) The binding energy of the system is the additional energy needed to move the particle to r → ∞ . This can be found by subtracting the total energy (-3 J) from the potential energy at r → ∞ (+1 J):

binding energy = 1 J - (-3 J) = 4 J