A 5.8 kg mass is attached to a spring with a stiffness of 490 N/m and is allowed to move along a friction-less horizontal surface. The mass is pulled back so that the spring is stretched 0.19 m from its equilibrium position. What will be the speed of the mass in m/s as it returns to the equilibrium position?

1.75 m/s

Explanation:

k = Spring constant = 490 N/m

m = Mass of object = 5.8 kg

x = Displacement of spring = 0.19 m

v = Speed of object at the equilibrium position

The potential energy of the spring will balance the kinetic energy of the mass

The speed of the mass as it returns to the equilibrium position is 1.75 m/s.

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