max Pe = max Ke
(1/2)k Xmax^2 = (1/2) m Vmax^2
(1/2)(400)(.1)^2 = (1/2)(.3)(Vmax)^2
40/3 = Vmax^2
The spring of force constant k = 400 N/m is set up horizontally with a 0.300 kg mass attached on it resting on a frictionless table. The mass is pulled so that the spring is stretched 0.100 m from the equilibrium point, and is then released from rest. Determine the speed of the system when the spring is at the equilibrium position.
1 answer
1 answer