A 1.00 -kg mass is suspended vertically from a spring with \(k=100 . \mathrm{N} / \mathrm{m}\) and oscillates with an amplitude of \(0.200 \mathrm{~m} .\) At the top of its oscillation, the mass is hit in such a way that it instantaneously moves down with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Determine a) its total mechanical energy, b) how fast it is moving as it crosses the equilibrium point, and c) its new amplitude.

First, let's find the initial kinetic energy (K.E) and potential energy (P.E) of the mass at the top of its oscillation. The initial kinetic energy is given by the formula: \(K.E = \cfrac{1}{2}mv^2\), where \(m\) is the mass, and \(v\) is the initial velocity. We're given the mass \(m=1.00 \text{ kg}\) and the initial velocity \(v=1.00 \text{ m/s}\). So, \(K.E = \cfrac{1}{2}(1.00)(1.00^2) = 0.50 \text{ J}\) (Joules) Now let's find the initial potential energy due to the spring. The potential energy is given by the formula: \(P.E = \cfrac{1}{2}kx^2\), where \(k\) is the spring constant and \(x\) is the displacement (amplitude). We're given the spring constant \(k = 100 \text{ N/m}\) and the amplitude \(x=0.200 \text{ m}\). So, \(P.E = \cfrac{1}{2}(100)(0.200^2) = 2.00 \text{ J}\)

Total mechanical energy (E) is the sum of kinetic energy and potential energy. So, \(E = K.E + P.E = 0.50 \text{ J} + 2.00 \text{ J} = 2.50 \text{ J}\) Therefore, the total mechanical energy of the mass is \(2.50 \text{ J}\).

At the equilibrium point, the potential energy is zero, so the total mechanical energy of the system is equal to the kinetic energy. Using the formula for kinetic energy \(K.E = \cfrac{1}{2}mv^2\) and our total mechanical energy, we can solve for the velocity at the equilibrium point: \(2.50 \text{ J} = \cfrac{1}{2}(1.00 \text{ kg})v^2\) \(v^2 = \cfrac{2.50*2}{1.00} \Rightarrow v = \sqrt{5.00}\) \(v \approx 2.24 \text{ m/s}\) So, the speed as it crosses the equilibrium point is approximately \(2.24 \text{ m/s}\).

To find the new amplitude, we use the conservation of mechanical energy. At the new maximum height (amplitude), the kinetic energy will be zero, and the total mechanical energy will be equal to the potential energy due to the spring: \(E = \cfrac{1}{2}kx^2\). We know the total mechanical energy (E) and the spring constant (k), so we can solve for the new amplitude (x): \(2.50 \text{ J} = \cfrac{1}{2}(100) x^2\) \(x^2 = \cfrac{2.50*2}{100} \Rightarrow x = \sqrt{0.0500}\) \(x \approx 0.224 \text{ m}\) So, the new amplitude of oscillation is approximately \(0.224 \text{ m}\). To summarize: a) The total mechanical energy is \(2.50 \text{ J}\). b) The speed as it crosses the equilibrium point is approximately \(2.24 \text{ m/s}\). c) The new amplitude is approximately \(0.224 \text{ m}\).