# Maximum acceleration of a spring

## Homework Statement

A spring with a spring constant of 5.0 N/m has a 0.25 kg box attached to one end such that the box is hanging down from the string at rest. The box is then pulled down another 14 cm from its rest position. Calculate the maximum height, the maximum speed, and the maximum acceleration of the box.

## The Attempt at a Solution

I found maximum speed and height but I cant find the maximum acceleration
ma= kx+mg
(o.25)a= (5)(0.14m)+(0.25)(9.81)
a=12.61 m's^2
what am I doing wrong?

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gneill
Mentor
Draw the FBD.

When the spring is stretched "another 14 cm", what is the total displacement of the spring? So what force does it apply? What direction?

Draw the FBD.

When the spring is stretched "another 14 cm", what is the total displacement of the spring? So what force does it apply? What direction?
is it ma=kx-mg instead of ma=kx+mg, even then I don't understand how acceleration can equal to 2.8m/s^2

gneill
Mentor
You didn't answer my question: When the spring is stretched "another 14 cm", what is the total displacement of the spring? What was the initial displacement when the mass was just hanging at rest on the end of the spring?