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Hooke's Law and Springs

  • Thread starter Gandor481
  • Start date
1. The problem statement, all variables and given/known data

Hooke's law describes a certain light spring of unstressed length 35.0 cm. When one end is attached to the top of a door frame and a 5.40 kg object is hung from the other end, the length of the spring is 42.00 cm.
(a) Find its spring constant.



(b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 150 N. Find the length of the spring in this situation.



2. Relevant equations

F=Kx

3. The attempt at a solution

A)
The spring constant is equal to
mg=kx
5.40*9.8=k*(.42-.35)
52.92=k(.07)
k=756 Nm

B)
∑F=150+150=300
300=kx
300/k=x
x=.39m

.39+.35=.74m

Which according to the book the correct answer is .54m which I was able to get if I only accounted for 150N being pulled instead of 300N since 150N is being pulled from both sides of the spring. I don't understand why it would only be 150N?
 

Simon Bridge

Science Advisor
Homework Helper
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Consider - when it was haging from the door, the weight pulled it by mg and the door pulled in the opposite direcetion by mg, yet you didn't say that kx=2mg did you? Why not?

Replace one of the people by a pole stuck in the ground - you get 150N one way due to the other person, and 150N the other way from the pole.

Note: force is a vector - so the total force on the spring is actually zero.
 

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