Question

Position when spring is unstrained NV Ху 5. In the figure above the mass attached to the spring is 0.2 kg. The spring's constant is 40 N/m The mass is stretched to a position of 0.3 m and released. The final position is 0.1m. 1. Find the change in potential energy of the system. 1 2. What is the work done by the spring?

Please show work and clearly mark answer.

Answer 1

Given: m = 0.2 kg k = 40 N/m x_0 = 0.3 m x_f = 0.1 m (1) potential energy of the system initially = 0.5 k * x_0^2 u_i = 0.5 times 40 times (0.3)^2 = 1.8 J final PE, of = 0.5 k * x_f^2 = 0.5 times 40 times (0.1)^2 = 0.2 J Hence, change in potential energy PE = Ui - Vf = 1.6 J \r\n (2) work done by the spring = change in energy of the spring = 1.6 J