Spring constant of a spring

  • Thread starter jannx3
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  • #1
jannx3
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Homework Statement


A mass of 50.0 g mass is hung from a spring, spring is elongated to 40 cm, an additional mass of 20g is added, with a new reading of 45cm

a) what is the spring constant for this spring?
b) another mass of 30g is added to the spring, what is the new reading?

Homework Equations



k= mg/x

The Attempt at a Solution


not sure if this is the right approach
a)
k= change in mass x g / change in height
0.020x9.8/0.05 =3.92n/m
b)mg=kx
0.03x9.8=3.92x
x=0.075 so it extend another 7.5cm?
 

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
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not sure if this is the right approach
a)
k= change in mass x g / change in height
0.020x9.8/0.05 =3.92n/m
b)mg=kx
0.03x9.8=3.92x
x=0.075 so it extend another 7.5cm?
Use a capital "N" for Newtons.
 
  • #3
22,428
5,269
The numbers in your solution don't seem to match the numbers in the problem statement.

Chet
 
  • #4
BiGyElLoWhAt
Gold Member
1,614
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why not just set up a N2L equation for this? If the spring stretches an amount x with a mass m attached to it and comes to rest again, then that means that net force is zero, and mg = kx, you only need 1 weight and 1 distance.
 
  • #5
Doc Al
Mentor
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The numbers in your solution don't seem to match the numbers in the problem statement.
They look fine to me.

A mass of 50.0 g mass is hung from a spring, spring is elongated to 40 cm,
I interpret the 40 cm as the total length of the spring, including its unstretched length.
 
  • #6
BiGyElLoWhAt
Gold Member
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132
Ahhh, I see what you're saying, so you do actually need two masses. Good call Doc.
elongated to 40 cm
 
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