Turning points occur where the kinetic energy is zero, so the total energy equals the potential energy: $$E = U(x)$$. We set $$4 = x^4 - 8x^2 + 12$$. This gives the equation $$x^4 - 8x^2 + 8 = 0$$. Let $$y = x^2$$, then $$y^2 - 8y + 8 = 0$$. Using the quadratic formula: $$y = \frac{8 \pm \sqrt{64 - 32}}{2} = \frac{8 \pm \sqrt{32}}{2} = 4 \pm 2\sqrt{2}$$. Since $$y = x^2 \geq 0$$, both solutions are valid: $$x^2 = 4 + 2\sqrt{2} \approx 6.83$$ or $$x^2 = 4 - 2\sqrt{2} = 4$$. The turning points are at $$x = \pm 2$$ m and $$x = \pm\sqrt{4 + 2\sqrt{2}}$$ m. Of the choices given, $$x = 2$$ m is a turning point.

The change in potential energy is the negative of the work done by the conservative force. $$U(r) - U(0) = -\int_0^r F_g(r') dr' = -\int_0^r (-G\frac{Mmr'}{R^3}) dr' = G\frac{Mm}{R^3} \int_0^r r' dr'$$. Evaluating the integral gives $$G\frac{Mm}{R^3} [\frac{1}{2}r'^2]_0^r = G\frac{Mm r^2}{2R^3}$$. Since $$U(0)=0$$, we have $$U(r) = \frac{GMmr^2}{2R^3}$$.

The gravitational potential energy of the system is given by $$U_g = -\frac{GMm}{r}$$. The initial potential energy at radius $$R$$ is $$U_i = -\frac{GMm}{R}$$. The final potential energy at radius $$3R$$ is $$U_f = -\frac{GMm}{3R}$$. The change in potential energy is $$\Delta U = U_f - U_i = -\frac{GMm}{3R} - (-\frac{GMm}{R}) = -\frac{GMm}{3R} + \frac{3GMm}{3R} = \frac{2GMm}{3R}$$. Since the result is positive, it is an increase.

The potential energy $$U(x)$$ is found by integrating the negative of the force: $$U(x) - U(0) = -\int_0^x F(x') dx'$$. With $$U(0)=0$$, we have $$U(x) = -\int_0^x (-kx' - \beta x'^3) dx' = \int_0^x (kx' + \beta x'^3) dx' = [\frac{1}{2}kx'^2 + \frac{1}{4}\beta x'^4]_0^x = \frac{1}{2}kx^2 + \frac{1}{4}\beta x^4$$.

The total potential energy is the sum of the elastic potential energy of the spring and the gravitational potential energy of the block-Earth system. The spring is stretched by a distance $$|y| = -y$$ (since y is negative). The elastic potential energy is $$U_s = \frac{1}{2}k(\text{stretch})^2 = \frac{1}{2}k(-y)^2 = \frac{1}{2}ky^2$$. The gravitational potential energy relative to $$y=0$$ is $$U_g = mgy$$. The total potential energy is $$U_{total} = U_s + U_g = \frac{1}{2}ky^2 + mgy$$.

When a spring of constant $$k$$ is cut in half, each half has a new spring constant $$k' = 2k$$. The original potential energy is $$U = \frac{1}{2}kx^2$$. We want the new potential energy $$U'$$ to be equal to $$U$$. Let the new stretch distance be $$x'$$. Then $$U' = \frac{1}{2}k'(x')^2 = \frac{1}{2}(2k)(x')^2 = k(x')^2$$. Setting $$U' = U$$ gives $$k(x')^2 = \frac{1}{2}kx^2$$. Solving for $$x'$$ yields $$(x')^2 = \frac{x^2}{2}$$, so $$x' = \frac{x}{\sqrt{2}}$$.

The work done by gravity is $$W_g = \int F_g dr$$. If the force $$F_g$$ is approximately constant over the displacement $$h$$, then $$W_g \approx -F_g h$$. Since $$W_g = -\Delta U_g$$, we get $$\Delta U_g \approx F_g h$$. Near the surface, $$F_g \approx mg$$, so $$\Delta U_g \approx mgh$$. The core of the approximation is that the gravitational field strength $$g$$, and thus the force, does not change significantly over the small height $$h$$.

The force vector is the negative gradient of the potential energy: $$\vec{F} = -\nabla U = -(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j})$$. First, find the partial derivatives: $$\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(3x^2y - 5y^3) = 6xy$$. $$\frac{\partial U}{\partial y} = \frac{\partial}{\partial y}(3x^2y - 5y^3) = 3x^2 - 15y^2$$. Now evaluate at the point (2, 1): $$\frac{\partial U}{\partial x}|{(2,1)} = 6(2)(1) = 12$$. $$\frac{\partial U}{\partial y}|{(2,1)} = 3(2^2) - 15(1^2) = 12 - 15 = -3$$. So, $$\vec{F} = -(12\hat{i} - 3\hat{j}) = -12\hat{i} + 3\hat{j}$$

This question tests understanding of elastic potential energy with a stiff spring and small compression. Elastic potential energy is calculated using U = ½kx², where the quadratic dependence on displacement means even small compressions in stiff springs can store significant energy. In this scenario, we have k = 500 N/m and x = 0.0800 m, giving U = ½(500)(0.0800)² = ½(500)(0.0064) = 1.60 J. Choice A is correct because it properly applies the formula with the squared displacement term. Choice B (3.20 J) is incorrect because it omits the ½ factor, a common error when students confuse the spring force formula with the energy formula. To help students: Emphasize the quadratic relationship between displacement and energy - doubling the compression quadruples the stored energy. Practice with various spring constants to build intuition about how stiffness affects energy storage.

This question tests understanding of elastic potential energy in spring systems, specifically calculating the energy stored when a spring is compressed. Elastic potential energy is given by the formula U = ½kx², where k is the spring constant and x is the displacement from equilibrium. In this scenario, we have a spring with k = 200 N/m compressed by x = 0.150 m, requiring us to substitute these values into the formula. Choice A is correct because U = ½(200)(0.150)² = ½(200)(0.0225) = 2.25 J. Choice B (4.50 J) is incorrect because it results from forgetting the ½ factor in the formula, a common mistake when students confuse the spring force formula (F = kx) with the energy formula. To help students: Emphasize that potential energy formulas often include a ½ factor due to integration, and practice dimensional analysis to verify that the units work out to joules. Create visual aids showing the parabolic relationship between compression distance and stored energy.