If the length of a #45 cm# spring increases to #86 cm# when a #1 kg# weight is hanging from it, what is the spring's constant?

Question

Elena Albarran · Accepted Answer

Approximately #24# newtons per meter

According to Hooke's law, the spring will lengthen in the direction that the weight is hanging from it. In this case, that would be: #F=kx# #F# is the force exerted in newtons #k# is the spring constant #x# is the extension of the spring The force here is the object's weight, which is #1 \ "kg"*9.8 \ "m/s"^2=9.8 \ "N"#. The spring constant is thus: #k=F/x# #=(9.8 \ "N")/(86 \ "cm"-45 \ "cm")# #=(9.8 \ "N")/(41 \ "cm")# #~~0.24 \ "N/cm"# #=24 \ "N/m"#

Axel Acevedo · Answer

undefined To find the spring constant, use Hooke's Law formula: $ F = k \cdot x $, where $ F $ is the force applied, $ k $ is the spring constant, and $ x $ is the displacement from the equilibrium position.

First, calculate the force applied by the weight: $ F = m \cdot g $, where $ m $ is the mass and $ g $ is the acceleration due to gravity (9.81 m/s$^2$).

$ F = 1 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 $

Then, calculate the change in length: $ \Delta x = x_{\text{final}} - x_{\text{initial}} $.

$ \Delta x = 86 \, \text{cm} - 45 \, \text{cm} $

Now, use Hooke's Law to find the spring constant: $ k = \frac{F}{\Delta x} $.

Substitute the values and convert units if necessary to find $ k $.