A horizontal spring with spring constant of \(9.82 \mathrm{N} / \mathrm{m}\) is attached to a block with a mass of \(1.24 \mathrm{kg}\) that sits on a frictionless surface. When the block is 0.345 m from its equilibrium position, it has a speed of \(0.543 \mathrm{m} / \mathrm{s}\) (a) What is the maximum displacement of the block from the equilibrium position? (b) What is the maximum speed of the block? (c) When the block is \(0.200 \mathrm{m}\) from the equilibrium position, what is its speed?

We are given the spring constant \((k = 9.82 \mathrm{N} / \mathrm{m})\), the mass of the block \((m = 1.24 \mathrm{kg})\), the initial speed \((v = 0.543 \mathrm{m} / \mathrm{s})\) when the block is at a displacement of \(x_1 = 0.345 \mathrm{m}\), and we need to find the maximum displacement \((x_{max})\), the maximum speed \((v_{max})\), and the speed \((v_2)\) when the block is \(x_{2=0.200 \mathrm{m}\) from the equilibrium position.

In this problem, we will assume the mechanical energy is conserved since the surface is frictionless. Therefore, we can write that the sum of kinetic energy and potential energy at any point in the motion remains constant throughout the motion: \(K_1 + U_1 = K_2 + U_2\) where \(K_1\) and \(U_1\) are the initial kinetic and potential energies, respectively, while \(K_2\) and \(U_2\) are the kinetic and potential energies at an arbitrary point in the motion.

At maximum displacement, the block will come to rest momentarily, implying that its kinetic energy is zero. Thus, we can rewrite the conservation of mechanical energy as: \(K_1 + U_1 = U_2\) and then solve for the maximum displacement \((x_{max})\): \(U_1 = \frac{1}{2}kx_1^2 = \frac{1}{2}(9.82)(0.345^2) = 0.582 \mathrm{J}\) \(K_1 = \frac{1}{2}mv^2 = \frac{1}{2}(1.24)(0.543^2) = 0.182 \mathrm{J}\) At maximum displacement, \(K_2 = 0\), thus: \(U_2 = \frac{1}{2}kx_{max}^2\) Equating \(K_1 + U_1\) to \(U_2\) and solving for \(x_{max}\): \(x_{max}^2 = \frac{2(K_1 + U_1)}{k} = \frac{2(0.182 + 0.582)}{9.82}\) \(x_{max} = \sqrt{0.154} = 0.392 \mathrm{m}\) So, the maximum displacement is \(0.392 \mathrm{m}\).

At maximum speed, the block will be at the equilibrium position, implying that its potential energy is zero. Thus, we can rewrite the conservation of mechanical energy as: \(K_1 + U_1 = K_2\) Solving for the maximum speed \((v_{max})\): \(K_2 = \frac{1}{2}mv_{max}^2\) Equating \(K_1 + U_1\) to \(K_2\) and solving for \(v_{max}\): \(v_{max}^2 = \frac{2(K_1 + U_1)}{m} = \frac{2(0.182 + 0.582)}{1.24}\) \(v_{max} = \sqrt{0.615} = 0.784 \mathrm{m} / \mathrm{s}\) So, the maximum speed is \(0.784 \mathrm{m} / \mathrm{s}\).

Since we know the sum of kinetic and potential energies is constant, we can rewrite the conservation of mechanical energy as \(K_1 + U_1 = K_2 + U_2\) where \(K_2\) and \(U_2\) are the kinetic and potential energy when the block is \(0.200 \mathrm{m}\) from the equilibrium position: \(U_2 = \frac{1}{2}kx_2^2 = \frac{1}{2}(9.82)(0.200^2) = 0.196 \mathrm{J}\) \(K_2 = K_1 + U_1 - U_2 = 0.182 + 0.582 - 0.196 = 0.568 \mathrm{J}\) \(v_2^2 = \frac{2K_2}{m} = \frac{2(0.568)}{1.24}\) \(v_2 = \sqrt{0.917} = 0.958 \mathrm{m} / \mathrm{s}\) So, the speed of the block when it is \(0.200 \mathrm{m}\) from the equilibrium position is \(0.958 \mathrm{m} / \mathrm{s}\). In summary: (a) The maximum displacement of the block is \(0.392 \mathrm{m}\). (b) The maximum speed of the block is \(0.784 \mathrm{m} / \mathrm{s}\). (c) The speed of the block when it is \(0.200 \mathrm{m}\) from the equilibrium position is \(0.958 \mathrm{m} / \mathrm{s}\).