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?

1 Answer
Jan 18, 2016

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#

Explanation:

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#.