Energy and SHO - physics

Question

A object with mass 0.248 kg is attached to a horizontal spring with a spring constant of 7.93 N/m. The object slides back and forth on a horizontal frictionless surface. The object is pulled back 8.50 cm and released.

What is the total mechanical energy of the object just as it is released?

How fast is the object traveling when it is a distance 5.01 cm from equilibrium (presume it is traveling in the positive direction).

Answer 1

Total ME = KE + EPE; KE = 0 at start, EKE = 1/2 k x² = 1/2 (7.93 N/m) (0.0850 m)² = 0.0286₄₇ N•m, so ME = 0.286 J at release.

At x = 5.01 cm, KE + 1/2 (7.93 N/m) (0.0501 m)² = 0.0286₅ J, KE = 1/2 m v² = 0.0286₅ - 0.00995 = 0.0187₀ J. v² = KE • 2/m = (0.0187₀ kg m²/s²) (2) /(0.248 kg) = 0.150₇ m²/s² and v = 0.388 m/s.