Potential Energy & Equilibrium - NEET Physics Chapterwise MCQs & PYQs

NEET Potential Energy & Equilibrium MCQs & PYQs

Practice NEET Potential Energy & Equilibrium Questions

Question 11:

moderate

The potential energy between two atoms, in a molecule, is given by \(U(x) = \frac{a}{x^{12}} – \frac{b}{x^6}\) where \(a\) and \(b\) are positive constants and \(x\) is the distance between the atoms. The atom is in stable equilibrium, when:

(1995)

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 \(x\).

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Question 12:

moderate

When a spring is subjected to \(4\text{ N}\) force its length is \(a\text{ metre}\). And if \(5\text{ N}\) is applied length is \(b\text{ metre}\). If \(9\text{ N}\) is applied length is:

(1999)

Concept: Hooke's Law. Formula: \(F = k(L - L_0)\), where \(L_0\) is original length. We have: (1) \(4 = k(a - L_0)\), (2) \(5 = k(b - L_0)\). From (1) and (2), we find \(L_0 = 5a - 4b\) and \(k = \frac{1}{b - a}\). For \(F=9\text{ N}\), \(9 = k(L_3 - L_0)\). Substitute \(k\) and \(L_0\): \(9 = \frac{1}{b - a}(L_3 - (5a - 4b))\). Solving for \(L_3\), we get \(L_3 = 9(b - a) + 5a - 4b = 9b - 9a + 5a - 4b = 5b - 4a\).

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Question 13:

moderate

Two similar springs \(P\) and \(Q\) have spring constants \(K_P\) and \(K_Q\) such that \(K_P > K_Q\). They stretched first by the same amount (case a), then by the same force (case b). The work done by the springs \(W_P\) and \(W_Q\) are related as in case (a) and case (b), respectively:

(2015)

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 \(W_P > W_Q\); \(W_Q > W_P\).

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Question 14:

moderate

A block of mass \(M\) is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant \(k\). The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be:

(2009)

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 \(x_{max} =\frac{2Mg}{k}\).

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Question 15:

moderate

A vertical spring with force constant \(k\) is fixed on a table. A ball of mass \(m\) at a height \(h\) above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance \(d\). The net work done in the process is:

(2007)

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 \(W_{net} = W_g + W_s = mg(h+d) - \frac{1}{2}kd^2\).

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Question 16:

moderate

Two springs \(A\) and \(B\) having spring constant \(K_A\) and \(K_B\) (\(K_A = 2K_B\)) are stretched by applying force of equal magnitude. If energy stored in spring \(A\) is \(E\) then energy stored in \(B\) will be:

(2001)

Concept: Energy stored in a spring under constant force. Formula: \(PE = \frac{F^2}{2K}\). When the same force \(F\) is applied, potential energy is inversely proportional to the spring constant (\(PE \propto 1/K\)). Given \(K_A = 2K_B\). The ratio \(\frac{PE_B}{PE_A} = \frac{K_A}{K_B}\). Substituting \(K_A = 2K_B\), we get \(\frac{PE_B}{E} = \frac{2K_B}{K_B} = 2\). Therefore, \(PE_B = 2E\).

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