Question

An ideal spring with a spring constant of \(15 \mathrm{N} / \mathrm{m}\) is suspended vertically. A body of mass \(0.60 \mathrm{kg}\) is attached to the unstretched spring and released. (a) What is the extension of the spring when the speed is a maximum? (b) What is the maximum speed?

Short answer

Answer: (a) The extension of the spring when the speed is maximum is 0.2024 m. (b) The maximum speed of the mass is 1.9524 m/s.

Step-by-step solution

Calculate gravitational force on the mass

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). \(F_g = (0.60 \mathrm{kg})(9.81 \mathrm{m/s^2}) = 5.886 \mathrm{N}\).

Find the extension at equilibrium

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: \(kx = F_g\) \(x = \frac{F_g}{k} = \frac{5.886 \mathrm{N}}{15 \mathrm{N/m}} = 0.3924 \mathrm{m}\).

Calculate the potential energy at equilibrium

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

Use conservation of energy to find maximum kinetic energy

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. \(K_{max} = U_s = 1.14508 \mathrm{J}\).

Calculate maximum speed of the mass

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

Calculate the extension when the speed is maximum

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 \(1.9524 \mathrm{m/s}\).