A mass of \(1.00 \mathrm{~kg}\) attached to a spring with a spring constant of \(100 .\) N/m oscillates horizontally on a smooth frictionless table with an amplitude of \(0.500 \mathrm{~m} .\) When the mass is \(0.250 \mathrm{~m}\) away from equilibrium, determine: a) its total mechanical energy; b) the system's potential energy and the mass's kinetic energy; c) the mass's kinetic energy when it is at the equilibrium point. d) Suppose there was friction between the mass and the table so that the amplitude was cut in half after some time. By what factor has the mass's maximum kinetic energy changed? e) By what factor has the maximum potential energy changed?

The total mechanical energy of the oscillating mass is the sum of its kinetic energy (KE) and potential energy (PE) due to the spring. In this case, the mechanical energy remains constant throughout the motion. To find the total mechanical energy, we first need to find the maximum potential energy when the mass is at its maximum amplitude, where the entire energy is potential energy and there's no kinetic energy. PE_max = (1/2) * k * A^2 Here, k is the spring constant, and A is the amplitude. PE_max = (1/2) * 100 * (0.500)^2 = 12.5 J Since the total mechanical energy is constant, we can say: E_total = PE_max = 12.5 J

At a distance of 0.250 m from the equilibrium point, the potential energy (PE) and kinetic energy (KE) can be calculated using Hooke's Law and the conservation of energy. PE = (1/2) * k * x^2 Here, x is the displacement from the equilibrium point. PE = (1/2) * 100 * (0.250)^2 = 3.125 J As the mechanical energy is conserved, the sum of PE and KE at this point must be equal to the total mechanical energy. KE = E_total - PE = 12.5 - 3.125 = 9.375 J

At the equilibrium point, the spring is at its natural length, and thus, there's no potential energy. Therefore, at this point, the entire mechanical energy is in the form of kinetic energy. KE_equilibrium = E_total = 12.5 J

If friction causes the amplitude to be halved after some time, we need to determine the new maximum potential energy and then find the maximum kinetic energy. New amplitude (A') = A / 2 = 0.500 / 2 = 0.250 m New maximum potential energy (PE_max') = (1/2) * k * (A')^2 = (1/2) * 100 * (0.250)^2 = 3.125 J The new total mechanical energy (E_total') = PE_max' = 3.125 J Now, at the maximum kinetic energy (occurring at the equilibrium point), there will be no potential energy. So, the new maximum kinetic energy is equal to the new total mechanical energy. KE_max' = E_total' = 3.125 J The factor by which the maximum kinetic energy has changed is: Factor = KE_max' / KE_equilibrium = 3.125 / 12.5 = 1 / 4 So, the maximum kinetic energy has changed by a factor of 1/4.

We found the new maximum potential energy (PE_max') to be 3.125 J. We can now find the factor by which the maximum potential energy has changed: Factor = PE_max' / PE_max = 3.125 / 12.5 = 1 / 4 So, the maximum potential energy has also changed by a factor of 1/4.