A weight of 0.1 kg was suspended from the spring, while the spring lengthened by 5 cm, then the weight was pulled

A weight of 0.1 kg was suspended from the spring, while the spring lengthened by 5 cm, then the weight was pulled back by 3 cm and released. Determine the speed at the moment of equilibrium passing: the total energy of the oscillating load.

h = 3 cm = 0.03 m;
m * g * h = 0.1 * 10 * 0.03 = 0.03 J;
m * v ^ 2/2 = 0.03 J;
v – speed at the moment of equilibrium passing.
v ^ 2 = 0.03 * 2 / 0.1 = 0.6 km / h;
v = √0.6 = 0.77 km / h.
Answer: 0.77 km / h.



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