A mass is vibrating at the end of force constant 225 N/m. The figure shows a graph of its position x as a function of time t. a) At what times between t = 0s and t = 4.5 s including the endpoints of the interval, is the mass not moving? b) How much energy

a) The mass is not moving when its position x is constant or not changing. Looking at the graph, we can see that the position x is constant between t = 1.5 s and t = 2.5 s, and also between t = 3.5 s and t = 4.5 s. Therefore, the mass is not moving during these time intervals. b) The total energy in a simple harmonic motion is given by the formula E = 0.5*k*A^2, where k is the force constant and A is the amplitude of the motion. From the graph, we can see that the amplitude is 0.08 m. Substituting k = 225 N/m and A = 0.08 m into the formula, we get E = 0.72 J. Therefore, the system originally contained 0.72 J of energy. c) The energy lost by the system can be calculated as the difference in energy between t = 1.00 s and t = 4.00 s. From the graph, we can see that the amplitude at t = 1.00 s is 0.08 m, and the amplitude at t = 4.00 s is \approx imately 0.03 m. Therefore, the energy at t = 1.00 s is E = 0.5*225*(0.08)^2 = 0.72 J, and the energy at t = 4.00 s is E = 0.5*225*(0.03)^2 = 0.30 J. The energy lost is then 0.72 J - 0.30 J = 0.42 J. d) The energy lost by the system is dissipated as heat due to friction or air resistance. For example, if there is a spring that is attached to the mass, then the energy could be lost due to friction between the spring and the mass as the mass oscillates back and forth. Alternatively, if there is air resistance, the energy could be lost as the mass pushes against air particles while moving.