If the length of a #24 cm# spring increases to #47 cm# when a #5 kg# weight is hanging from it, what is the spring's constant?

Question

Lily Adams · Accepted Answer

Better to rewrite the question in metres: "#0.24 m# spring increases #0.47 m#..." Spring constant, #k = F/(Delta d) = (m*g)/(Delta d) = (5*9.8)/(0.47-0.24) = 21.3 Nm^-1# The spring constant, #k#, of a spring is expressed in newton per metre #(Nm^-1)#: for each additional #N# of force the spring expands by #k (m)#. It's important to use correct SI units for the quantities: centimetre is not a base SI unit, metre is, so convert the #cm# to #m#. I used #Delta d# to represent the change in length of the spring - final length of #0.47 m# minus initial length #0.24 m#. The force acting on the spring is the weight force of the mass, which is its mass, #5 kg# times #g = 9.8 Nkg^-1# (more often written as #ms^-2#, but this is an exactly equivalent unit that makes more sense in this context). Calculating all this leads to a spring constant equal to #21.3 Nm^-1#.

Lily Adams · Answer

undefined The spring constant, $ k $, can be calculated using Hooke's Law:

$$ F = -kx $$

Where:
- $ F $ is the force applied to the spring (weight of the object),
- $ k $ is the spring constant,
- $ x $ is the displacement from the equilibrium position.

Given:
- $ F = mg $ (weight of the object),
- $ m = 5 $ kg,
- $ g = 9.8 $ m/s$^2$ (acceleration due to gravity),
- Initial length ($ x_1 $) = 24 cm = 0.24 m,
- Final length ($ x_2 $) = 47 cm = 0.47 m.

Using Hooke's Law:

$$ F = k(x_2 - x_1) $$

$$ mg = k(x_2 - x_1) $$

$$ k = \frac{mg}{x_2 - x_1} $$

$$ k = \frac{5 \times 9.8}{0.47 - 0.24} $$

$$ k ≈ \frac{49}{0.23} $$

$$ k ≈ 213.04 \, \text{N/m} $$

So, the spring constant is approximately $ 213.04 \, \text{N/m} $.