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)

a=.21

how is this wrong

Answer 1

If the spring is stretched 18cm and the spring constant is 65N/m, then the force is 0.018 * 65

The object will accelerate in accord with F = m*a

0.018 * 65 = 5.5 * a

a = 0.018 * 65 / 5.5 = 0.213m/sec^2

I don't know where you got the equations you used.....

One tip: include the units to make sure you have the numbers arranged correctly:

Force: F = 0.018 meters * 65 Newtons/meter = ____Newtons

F = ma:

Newtons = kg * meters/sec^2 (the definition of a Newton)