The conservative force is related to the potential energy by $$F(x) = -\frac{dU}{dx}$$. Differentiating $$U(x)$$ gives $$\frac{dU}{dx} = -12x^2 + 12x$$. Therefore, the force is $$F(x) = -(-12x^2 + 12x) = 12x^2 - 12x$$. Evaluating at $$x = -1$$ m gives $$F(-1) = 12(-1)^2 - 12(-1) = 12(1) + 12 = 24$$ N.

Work is defined by the dot product of force and displacement, $$W = \int \vec{F} \cdot d\vec{s}$$. The normal force is, by definition, always perpendicular to the surface of the track. The instantaneous displacement (and thus velocity) of the car is always tangent to the track. Since the normal force and displacement vectors are always perpendicular, their dot product is zero, and the normal force does no work.

To remain in contact at the top of the loop (height $$2R$$), the centripetal force must at least equal the gravitational force, so $$mg = mv_{top}^2/R$$, which gives the minimum speed $$v_{top}^2 = gR$$. By conservation of energy from the start to the top of the loop: $$mgH = mg(2R) + \frac{1}{2}mv_{top}^2$$. Substituting $$v_{top}^2$$: $$mgH = 2mgR + \frac{1}{2}mgR = 2.5mgR$$. Thus, $$H = 2.5R$$.

Air resistance is a non-conservative force that does negative work on the ball as it moves upward, dissipating mechanical energy. The change in mechanical energy is equal to the work done by non-conservative forces: $$W_{nc} = E_f - E_i$$. Here, $$E_i = K_0$$ and $$E_f = U_{max}$$. Since $$W_{air} < 0$$, we have $$U_{max} - K_0 < 0$$, which implies $$K_0 > U_{max}$$.

First, use conservation of momentum for the perfectly inelastic collision: $$mv = (M+m)V$$, where $$V$$ is the speed just after impact. So $$V = \frac{mv}{M+m}$$. Then, use conservation of mechanical energy for the swing: $$\frac{1}{2}(M+m)V^2 = (M+m)gh$$. Solving for $$h$$ gives $$h = \frac{V^2}{2g}$$. Substituting the expression for $$V$$ gives $$h = \frac{1}{2g} (\frac{mv}{M+m})^2 = \frac{m^2 v^2}{2g(M+m)^2}$$.

The total energy is constant and equal to the initial potential energy: $$E_{total} = \frac{1}{2}kA^2$$. We want the position $$x$$ where kinetic energy $$K$$ equals potential energy $$U$$. At this point, $$E_{total} = K + U = U + U = 2U$$. So, $$\frac{1}{2}kA^2 = 2(\frac{1}{2}kx^2) = kx^2$$. Solving for $$x$$ gives $$x^2 = A^2/2$$, so $$x = A/\sqrt{2}$$.

By conservation of mechanical energy, the initial potential energy $$U_i = mgH$$ (relative to the lowest point) is fully converted to kinetic energy $$K_f$$ at the lowest point, since $$U_f = 0$$ and the bob starts from rest ($$K_i = 0$$). Thus, $$E_i = E_f$$ implies $$mgH = K_f$$.

On the incline, conservation of energy gives the block's kinetic energy at the bottom as $$K = mgh$$. On the horizontal surface, the work done by friction, $$W_f = -f_k d = -\mu_k mgd$$, must equal the change in kinetic energy, which is $$0 - K = -mgh$$. Setting $$-\mu_k mgd = -mgh$$ and solving for $$d$$ gives $$d = h/\mu_k$$.

The total mechanical energy of the system, set at the moment of release, is $$E = \frac{1}{2}kA^2$$. At any position $$x$$, the energy is $$E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$$. By conservation of energy, $$\frac{1}{2}kA^2 = \frac{1}{2}k(A/2)^2 + \frac{1}{2}mv^2$$. This simplifies to $$\frac{1}{2}kA^2 = \frac{1}{8}kA^2 + \frac{1}{2}mv^2$$, which gives $$\frac{3}{8}kA^2 = \frac{1}{2}mv^2$$. Solving for $$v$$ yields $$v = \sqrt{\frac{3kA^2}{4m}}$$.

According to the work-energy theorem, the work done on the object equals its change in kinetic energy. Since it starts from rest, its final kinetic energy is equal to the work done. The work is calculated by integrating the force over the displacement: $$W = \int_{0}^{2} F(x) dx = \int_{0}^{2} 6x^2 dx = [2x^3]_0^2 = 2(2)^3 - 0 = 16$$ J.