A single mass (m1 = 3.1 kg) hangs from a spring in a motionless elevator. The spring constant is k = 338 N/m.

Question

1. What is the distance the spring is stretched from its unstretched length?
2. Now, three masses (m1 = 3.1 kg, m2 = 9.3 kg and m3 = 6.2) hang from three identical springs in a motionless elevator. The springs all have the same spring constant given above.

What is the magnitude of the force the bottom spring exerts on the lower mass?
3. What is the distance the middle spring is stretched from its equilibrium length?

Answer 1

1. Given that known variable k = 338 N/m and F=mg=6.2(9.8), we will use Hooke's Law(F=-kx) to calculate the stretched distance for the spring. F=mg=-kx which is 60.76=338x. Therefore, x = 0.18m

Answer 2

2. F=ma=6.2(9.8)=60.76N
*If you are calculating with an additional acceleration, you will add that to g.

Answer 3

F=-kx=ma
m=m2+m3=9.3+6.2
a = g+elevator acceleration = -9.8
k=338
-9.8(15.5)=-338x
x=0.449m