91Ó°ÊÓ

To calculate the gravitational force acting on the mass, we use the formula: \(F_g = mg\). Here, m = 0.60 kg and \(g = 9.81 \mathrm{m/s^2}\) (acceleration due to gravity). \(F_g = (0.60 \mathrm{kg})(9.81 \mathrm{m/s^2}) = 5.886 \mathrm{N}\).

When the mass is at equilibrium, the gravitational force will be counterbalanced by the spring force. So, \(F_s = F_g\). Using Hooke's law, we get: \(kx = F_g\) \(x = \frac{F_g}{k} = \frac{5.886 \mathrm{N}}{15 \mathrm{N/m}} = 0.3924 \mathrm{m}\).

The potential energy stored in the spring at equilibrium extension is given by: \(U_s = \frac{1}{2}kx^2\) \(U_s = \frac{1}{2}(15 \mathrm{N/m})(0.3924 \mathrm{m})^2 = 1.14508 \mathrm{J}\).

When the speed of the mass is at its maximum, the potential energy stored in the spring will be a minimum. Thus, the conservation of energy implies that the total energy conserved in the system will be equal to the maximum kinetic energy, \(K_{max}\) of the mass. \(K_{max} = U_s = 1.14508 \mathrm{J}\).

Use the formula for kinetic energy, \(K = \frac{1}{2}mv^2\), to find the maximum speed, \(v_{max}\): \(v_{max}^2 = \frac{2K_{max}}{m}\) \(v_{max}= \sqrt{\frac{2(1.14508 \mathrm{J})}{0.60 \mathrm{kg}}} = \sqrt{3.81693}\) \(v_{max} = 1.9524 \mathrm{m/s}\).

Since the kinetic energy of the mass is maximum, the potential energy of the spring will be minimum. As potential energy is given by \(U_s = \frac{1}{2}kx^2\), this occurs when the square of the extension is the smallest possible value. At this moment, we know the total mechanical energy \(E\) is equal to \(K_{max}\). Using conservation of energy, we can write: \(E=K+U_s\): \(1.14508 \mathrm{J} = \frac{1}{2}(0.60 \mathrm{kg})(1.9524 \mathrm{m/s})^2 + \frac{1}{2}(15 \mathrm{N/m})x^2\) Solving for x: \(x^2 = \frac{1.14508 - (0.60)(1.9524)^2 / 2}{15 / 2}\) \(x^2 = 0.0409\) \(x = \sqrt{0.0409} = 0.2024 \mathrm{m}\) Therefore, (a) the extension of the spring when the speed is maximum is \(0.2024 \mathrm{m}\) and (b) the maximum speed of the mass is \(1.9524 \mathrm{m/s}\).