# A mass hung from two attached springs

shawli

## Homework Statement

A light spring with constant k1 is hung from an elevated support. From its lower end a second light spring is hung, which has spring constant k2. An object of mass m is hung at rest from the lower end of the second spring.

a) Find the total extension distance of the pair of the springs.

F= ma
Fs = kx

## The Attempt at a Solution

I treated the two springs as a single device and made the following equation for the forces acting on the mass:

Fnet = 0
Fs1 + Fs2 - mg = 0
k1x + k2x = mg
(k1 + k2)x = mg
x = mg/(k1 + k2)

But this answer is incorrect. (Correct answer: x = mg(1/k1 + 1/k2).

Would someone be able to clarify what is happening in this system? I seem to be missing something...

## Answers and Replies

carhartt
Seems like you need to treat each spring as a separate system. They have different spring constants so will put out different forces.

k1x = mg gives x1=mg/k1 similarly x2 = mg/k2
add x1 and x2 gives you total distance the spring is stretched.

Just my \$0.02, this is my first year taking calculus based physics as well.

shawli
Hm I see. I was wrong to assume the forces on each spring would be the same...

Thank you! And I hope your physics course(s) are going well :)