Maximum acceleration of a spring

AI Thread Summary
The discussion revolves around calculating the maximum acceleration of a box attached to a spring with a spring constant of 5.0 N/m. The box, weighing 0.25 kg, is initially at rest and then pulled down an additional 14 cm. The user initially calculates maximum acceleration incorrectly using the formula ma = kx + mg, leading to confusion over the correct approach. Clarification is sought on the total displacement of the spring and the forces acting on the box, emphasizing the need to consider both the spring force and gravitational force. The correct formula should reflect the net force acting on the box, which is crucial for accurately determining the maximum acceleration.
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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.

Homework Equations

The Attempt at a Solution


I found maximum speed and height but I can't find the maximum acceleration
ma= kx+mg
(o.25)a= (5)(0.14m)+(0.25)(9.81)
a=12.61 m's^2
the answer is 2.8m/s^2
what am I doing wrong?
 
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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?
 
gneill said:
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
 
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?
 

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