Calculating Spring Stretch in Series: 2.8 N/cm and 6 N/cm

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SUMMARY

The discussion focuses on calculating the stretch of two springs in series with constants of 2.8 N/cm and 6 N/cm, supporting a mass of 41 N. The correct method involves using Hooke's Law, where the force applied to a spring is equal to the spring constant multiplied by the distance stretched. The user initially miscalculated by dividing 41 N by 2.8 N/cm instead of using the correct units, leading to confusion in determining the stretch distances for both springs.

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muna580
I have this homework question which I don't understand how to do it.

Since I am not homew right now, I can't post up the picture. I will explain what the picture looks like.

Picture:

___ = Wall
$ = Spring
| = Connection
(*) = mass
____________
$ 2.8N/cm
$ 6N/cm
(*) 41

So basically we have 2 springs with diffrent elasticities hanging from the wall. And below the springs is a mass with 41N.

1. What distance will the spring of constant 2.8 N/cm stretch? Answer in units of cm.

2. In the same series spring system, what distance will the spring of constant 6 N/cm stretch? Answer in units of cm.
 
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muna580 said:
I have this homework question which I don't understand how to do it.

Since I am not homew right now, I can't post up the picture. I will explain what the picture looks like.

Picture:

___ = Wall
$ = Spring
| = Connection
(*) = mass
____________
$ 2.8N/cm
$ 6N/cm
(*) 41

So basically we have 2 springs with diffrent elasticities hanging from the wall. And below the springs is a mass with 41N.

1. What distance will the spring of constant 2.8 N/cm stretch? Answer in units of cm.

2. In the same series spring system, what distance will the spring of constant 6 N/cm stretch? Answer in units of cm.
Assuming the springs are massless, how much weight must each spring support?
 
Oh wait, I got the answer. I was doing the calcuations wrong. Instead of deviding 41N by the 2.8 N/cm, I was deviding .41 by 2.8. I was trying to convert 41N to cm or something like that. lol
 

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