Maximum acceleration of a spring

  • Thread starter rrosa522
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  • #1
rrosa522
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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?
 

Answers and Replies

  • #2
gneill
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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?
 
  • #3
rrosa522
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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?
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
gneill
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2,886
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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