a) The maximum speed of the object at x = 6a so that it is confined to the region 4a < x < 8a is the speed at the point where the potential energy curve intersects the horizontal lines representing the energy levels of 16.77 J and 5.59 J. This occurs at x = 5a and x = 7a, respectively. The change in potential energy between these points is:
ΔU = U(5a) - U(7a) = (3)(5.59 J) = 16.77 J
Using the conservation of energy principle, the maximum kinetic energy and speed of the object at x = 6a is:
KE = ΔU = 16.77 J
(1/2)mv^2 = 16.77 J
v = sqrt((2*16.77 J)/(0.35 kg)) = 8.19 m/s
b) The maximum speed of the object at x = 6a so that it is confined to the region 3a < x < 9a is the speed at the point where the potential energy curve intersects the horizontal lines representing the energy levels of 16.77 J and 0 J. This occurs at x = 3.5a and x = 8.5a, respectively. The change in potential energy between these points is:
ΔU = U(3.5a) - U(8.5a) = (6)(5.59 J) = 33.54 J
Using the conservation of energy principle, the maximum kinetic energy and speed of the object at x = 6a is:
KE = ΔU = 33.54 J
(1/2)mv^2 = 33.54 J
v = sqrt((2*33.54 J)/(0.35 kg)) = 12.36 m/s
c) The change in potential energy between x = 6a and x = 0 is:
ΔU = U(6a) - U(0) = (10)(5.59 J) = 55.9 J
Let v' be the speed of the object when it is at x = 0. The conservation of energy principle gives:
(1/2)mv^2 - ΔU = (1/2)mv'^2
(1/2)(0.35 kg)(1.172v)^2 - 55.9 J = (1/2)(0.35 kg)(v')^2
Solving for v', we get:
v' = sqrt[((1/2)(0.35 kg)(1.172v)^2 - 55.9 J)/(0.5*0.35 kg)] = 6.05 m/s