spring acceleration

Question

An object of mass 5.50 kg attached to an ideal massless spring is pulled with a steady horizontal force across a frictionless level surface. If the spring constant is 65.0 N/m and the spring is stretched by 18.0 cm , what is the magnitude of the acceleration of the object?

a=Aω^2

ω=√(k/m)

a=(.018)(√(65/5.5))²

a=.21

how is this wrong

Answer 1

Calculate the spring force. That will be the same as the force applied on the object. Then divide by the mass of the object to get its acceleration.

F = kx = ma

a = kx/m = (65.0 N/m)(0.18 m) / (5.50 kg) = 2.13 m/s²

It looks like your error was in converting the 18.0 cm to meters.