Question 31:
easyAssertion (A): Frictional forces are conservative forces.
Reason (R): Potential energy can be associated with frictional forces.
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.
Question 32:
easyState the incorrect statement among the following
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.
Question 33:
moderateThe potential energy of particle in a force field is \(U = \frac{A}{r} – \frac{B}{r^2}\) where \(A\) and \(B\) are positive constants and \(r\) is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is:
(2012 Pre)
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 \(r = \frac{2A}{B}\).
Question 34:
easyThe potential energy of a system increases if work is done:
(2011 Pre)
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).
Question 35:
moderateThe 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 \(x\).
Question 36:
moderateTwo 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 \(W_P > W_Q\); \(W_Q > W_P\).
Question 37:
moderateA 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 \(x_{max} =\frac{2Mg}{k}\).
Question 38:
moderateA 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 \(W_{net} = W_g + W_s = mg(h+d) - \frac{1}{2}kd^2\).
Question 39:
easyThe potential energy of a long spring when stretched by \(2\text{ cm}\) is \(U\). If the spring is stretched by \(8\text{ cm}\) the potential energy stored in it is:
(2006)
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, \(PE_2 = 16U\).
Question 40:
easyWhen a long spring is stretched by \(2\text{ cm}\), its potential energy is \(U\). If the spring is stretched by \(10\text{ cm}\), the potential energy stored in it will be:
(2003)
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, \(PE_2 = 25U\).