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)
1. \(4b - 3a\)
2. \(5b - a\)
3. \(5b - 4a\)
4. \(5b - 2a\)
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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 \(L_3 = 9(b - a) + 5a - 4b = 9b - 9a + 5a - 4b = 5b - 4a\).
The kinetic energy acquired by a mass \(m\) in travelling distance \(d\), starting from rest, under the action of a constant force is directly proportional to:
(1994)
1. \(m\)
2. \(m^0\)
3. \(\sqrt{m}\)
4. \(1/\sqrt{m}\)
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Concept: Work-Energy Theorem. Formula: \(W = Fd = \Delta KE\). Starting from rest, \(KE_i = 0\). So, the final kinetic energy \(KE_f = Fd\). If the force \(F\) and distance \(d\) are constant, the work done \(Fd\) is constant. Therefore, the kinetic energy acquired is independent of mass \(m\), meaning it is proportional to \(m^0\).
A particle of mass \(M\) is moving in a horizontal circle of radius \(R\) with uniform speed \(v\). When it moves from one point to a diametrically opposite point, its:
(1992)
1. Kinetic energy change by \(Mv^2/4\)
2. Momentum does not change
3. Momentum change by \(2Mv\)
4. Kinetic energy changes by \(Mv^2\)
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Concept: Momentum and kinetic energy in uniform circular motion. Formula: Momentum \(p = Mv\), Kinetic Energy \(KE = \frac{1}{2}Mv^2\). Since speed \(v\) is uniform, KE remains constant (\(\Delta KE = 0\)). At diametrically opposite points, the direction of velocity reverses. If initial momentum is \(\vec{p_1} = M\vec{v}\), then final momentum is \(\vec{p_2} = -M\vec{v}\). The change in momentum is \(\Delta\vec{p} = \vec{p_2} - \vec{p_1} = -2M\vec{v}\). The magnitude of the change is \(2Mv\).
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)
1. \(W_P = W_Q\); \(W_P = W_Q\)
2. \(W_P > W_Q\); \(W_Q > W_P\)
3. \(W_P < W_Q\); \(W_Q < W_P\)
4. \(W_P = W_Q\); \(W_P > W_Q\)
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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 \(W_P > W_Q\); \(W_Q > W_P\).
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)
1. \(2Mg/k\)
2. \(4Mg/k\)
3. \(Mg/2k\)
4. \(Mg/k\)
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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 \(x_{max} =\frac{2Mg}{k}\).
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)
1. \(mg(h+d) - \frac{1}{2}kd^2\)
2. \(mg(h-d) - \frac{1}{2}kd^2\)
3. \(mg(h-d) + \frac{1}{2}kd^2\)
4. \(mg(h+d) + \frac{1}{2}kd^2\)
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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 \(W_{net} = W_g + W_s = mg(h+d) - \frac{1}{2}kd^2\).
When 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)
1. \(U/5\)
2. \(5U\)
3. \(10U\)
4. \(25U\)
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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, \(PE_2 = 25U\).