# Spring constant of a spring

jannx3

## 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?

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?

Staff Emeritus
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.

1 person
Mentor
The numbers in your solution don't seem to match the numbers in the problem statement.

Chet

Gold Member
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.

Mentor
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.

Gold Member
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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