Question

A single mass m1 = 3.6 kg hangs from a spring in a motionless elevator. The spring is extended x = 15.0 cm from its unstretched length.

Now, three masses m1 = 3.6 kg, m2 = 10.8 kg and m3 = 7.2 kg hang from three identical springs in a motionless elevator. The springs all have the same spring constant that you just calculated above.

What is the distance the lower spring is stretched from its equilibrium length?

Answer 1

Okay so we need to use hooke's law:
\[F = -kx\]
In this case \(x\) is how far it pulls it down and \(F\) is the force the weight pulls downward due to gravity. We use this formula to find \(k\), which is the spring constant.
So we also need to use Newton's laws of motion:
\[F = ma\]
In this case, we have \(m\) as the mass of the weight and \(a\) as the gravitational acceleration constant \(9.82m/s^2\).

Do you think you can do it?

Answer 2

i get 2.35 but it tells me that its wrong and im making a power of ten error?

Answer 3

what is 3?

Answer 4

it says thats wrong

Answer 5

what did u get as the spring constant?

Answer 6

use meters, not cm.

Answer 7

oh yes thanks!blunder!!!!!

Answer 8

ok great got it thank u

Answer 9

What is the distance the lower spring is stretched from its equilibrium length? what about that question?