If the length of a #32 cm# spring increases to #53 cm# when a #15 kg# weight is hanging from it, what is the spring's constant?

1 Answer
Feb 2, 2016

Answer:

700 N/m

Explanation:

The calculation is based on Hooke's Law and is only applicable for simple springs where the deflection or compression is not excessive. In equation form it is expressed as F= ky.

Where F is the applied force in units of Newtons. K is the spring constant and y the deflection or compression in metres. As there is a mass attached to the spring there is a deflection of 0.21 m. The vertical force can be calculated using Newtons second Law as F = ma. Where m is the objects mass in kilograms and a the gravitational acceleration (9.8 m/s^2)

To confirm whether Hooke's law is valid, you could plot a graph of the applied force F against the deflection y for several weights. If the graph is linear you can safely assume Hookes law is valid. The slope of the graph will be the spring constant K.