# Maximum acceleration of a spring

1. Apr 12, 2016

### rrosa522

1. The problem statement, all variables and given/known data
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.
2. Relevant equations

3. 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?

2. Apr 12, 2016

### Staff: 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?

3. Apr 12, 2016

### rrosa522

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

4. Apr 12, 2016

### Staff: 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?