Question

A spring hanging from a hook has a relaxed length of 8.1 cm with no mass attached to it. When a 0.41 kg mass is hung from the spring the spring stretches to a new equilibrium length of 12 cm. The mass is then pulled downward an additional 5.8 cm from the equilibrium position and released from rest. How fast (in m/s) is the mass moving when it first passes through the original equilibrium position of the spring (the relaxed length)? State the numerical value of the speed (in m/s) only, without units. Type answer here. Be attentive to spelling.

Answer 1

For extension in spring when m = 0.41 kg mass is attached mg = kx_0^' 0.41 times 9.81 = k (3.9 times 10^-2 m) k = 103.13 N/m Now when it again stretches x = 5.8 cm Then for spring & mass system mechanical energy remains conserved. (KE)_mass + [(P.E)_spring + (PE)_mass] = conct. [1/2 m v_1^2 + 1/2 k (x + x_0^')^2 + mgh]_before = [0]_after [0 + 1/2 times 103.13 times (9.7 cm)^2 + 0] = 1/2 mv^2 + 0 + mg (9.7 cm) 1/2 times 103.13 times (9.7 times 10^-2)^2 = 1/2 times (.41) times v^2 + 0.41 times 9.81 times 9.7 times 10^-2 485 = 0.205 v^2 + 0.39 V = 0.68 m/sec Ans.