An elastic cord is 61 cm long when a weight of 75 N hangs from it but is 85 cm long when a weight of 210 N hangs from it. What is the "spring" constant #k# of this elastic cord?

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Answer 1 · 2017-03-11T09:11:24

The spring constant is #=577Nm^-1#

Let the length of the elastic cord at rest #=l m#

Therefore,

In the first case, #k=F_1/(Deltax_1)=75/(0.61-l)#

In the second case, #k=F_2/(Deltax_2)=210/(0.85-l)#

We have 2 equations with 2 unknowns.

#75/(0.61-l)=210/(0.85-l)#

#75(0.85-l)=210(0.61-l)#

#210l-75l=210*0.61-75*0.85#

#135l=64.35#

#l=64.35/135=0.48#

So,

#k=75/(0.61-0.48)=576.9Nm^-1#