If the length of a #21 cm# spring increases to #45 cm# when a #2 kg# weight is hanging from it, what is the spring's constant?

1 Answer
Jan 8, 2017

Answer:

I got #k=81.7N/m#

Explanation:

We can us Hooke's Law to relate the elastic force of a spring of elastic constant #k# after being elongated of #x#:
#F_(el)=-kx#
Also, Newton's Second Law will help us to consider the forces involved through:
#SigmavecF=m veca#

let us see:
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So at the equilibrium the elastic force and the weight (#W=mg#) will give zero acceleration:
#-kx-mg=0#
But the elongation is directed downwards so we write, using numbers:
#-k(-0.24)-(2*9.8)=0#
So:
#k=(2*9.8)/0.24=81.7N/m#
I changed the lengths into meters because I am used to this but if you need it in N/cm use #24cm# instead of #0.24m# in the above.