Frictional forces are non-conservative forces because the work done by them depends on the path taken and energy is dissipated as heat. Potential energy can only be associated with conservative forces (e.g., gravitational, elastic).

Therefore, both assertion and reason are false.

In the presence of non-conservative internal forces (like friction or impact forces in inelastic collisions), total kinetic energy is not conserved, making statement (4) incorrect.

For equilibrium, the force is zero: \(F = -\frac{dU}{dr} = 0\). Given \(U = \frac{A}{r} - \frac{B}{r^2}\), if the question implicitly means \(U = \frac{A}{r^2} - \frac{B}{r}\), then \(\frac{dU}{dr} = -\frac{2A}{r^3} + \frac{B}{r^2}\). Setting \(\frac{dU}{dr} = 0\) gives \(\frac{2A}{r^3} = \frac{B}{r^2} \Rightarrow r = \frac{2A}{B}\). Checking stability, \(\frac{d^2U}{dr^2} = \frac{6A}{r^4} - \frac{2B}{r^3}\), which is positive at \(r = \frac{2A}{B}\).

Potential energy is associated with conservative forces. Work done by a conservative force is \(W_c = -\Delta U\). Therefore, if potential energy increases (\(\Delta U > 0\)\), then \(W_c\) must be negative. This happens when work is done by the system against a conservative force (e.g., lifting an object against gravity).

For equilibrium, the force is zero: \(F = -\frac{dU}{dx} = 0\). Given \(U(x) = ax^{-12} - bx^{-6}\), then \(\frac{dU}{dx} = -12ax^{-13} + 6bx^{-7}\). Setting \(\frac{dU}{dx} = 0\) gives \(\frac{12a}{x^{13}} = \frac{6b}{x^7} \Rightarrow 12a = 6bx^6 \Rightarrow x^6 = \frac{12a}{6b} = \frac{2a}{b}\). So, \(x = (\frac{2a}{b})^{1/6}\). For stable equilibrium, \(\frac{d^2U}{dx^2} > 0\), which holds true for this value of \(x\).

Concept: Work done to stretch a spring. Formula: \(W = \frac{1}{2}Kx^2\) and \(W = \frac{F^2}{2K}\). Given \(K_P > K_Q\). Case (a): Same extension \(x\). \(W propto K\), so \(W_P > W_Q\). Case (b): Same force \(F\). \(W \propto 1/K\), so \(W_P W_P\). Combining these, the correct option is \(W_P > W_Q\); \(W_Q > W_P\).

Concept: Conservation of Mechanical Energy. Initial state: \(KE_i = 0\), \(PE_i = 0\) (reference at initial position). Final state (maximum extension \(x_{max}\)): \(KE_f = 0\), \(PE_f = -Mgx_{max} + frac{1}{2}kx_{max}^2\). By energy conservation, \(KE_i + PE_i = KE_f + PE_f\), so \(0 = -Mgx_{max} + \frac{1}{2}kx_{max}^2\). Solving for \(x_{max}\), we get \(Mg = \frac{1}{2}kx_{max}\), which yields \(x_{max} =\frac{2Mg}{k}\).

Concept: Work done by conservative forces. Formula: \(W_g = mg\Delta h\), \(W_s = -\frac{1}{2}kx^2\).

The total vertical distance the mass falls is \(h+d\), so work done by gravity is \(W_g = mg(h+d)\). The spring is compressed by \(d\), so work done by the spring is \(W_s = -\frac{1}{2}kd^2\). The net work done by these forces is \(W_{net} = W_g + W_s = mg(h+d) - \frac{1}{2}kd^2\).

Concept: Potential energy stored in a spring. Formula: \(PE = \frac{1}{2}kx^2\). Potential energy is proportional to the square of the extension (\(PE \propto x^2\)). Given \(PE_1 = U\) for \(x_1 = 2\text{ cm}\). We need \(PE_2\) for \(x_2 = 8\text{ cm}\). \(\frac{PE_2}{PE_1} = \left(\frac{x_2}{x_1}\right)^2 = \left(\frac{8\text{ cm}}{2\text{ cm}}\right)^2 = (4)^2 = 16\). Therefore, \(PE_2 = 16U\).

Concept: Potential energy stored in a spring. Formula: \(PE = \frac{1}{2}kx^2\). Potential energy is proportional to the square of the extension (\(PE \propto x^2\)). Given \(PE_1 = U\) for \(x_1 = 2\text{ cm}\). We need \(PE_2\) for \(x_2 = 10\text{ cm}\). \(\frac{PE_2}{PE_1} = \left(\frac{x_2}{x_1}\right)^2 = \left(\frac{10\text{ cm}}{2\text{ cm}}\right)^2 = (5)^2 = 25\). Thus, \(PE_2 = 25U\).