If the length of a $27 cm$ spring increases to $45 cm$ when a $4 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2017-05-21T07:33:03

The spring constant is $=217.8kgs^-2$

The equation of the elongation of the spring is

$F=k*Deltax$

The force is $F=4gN$

The elongation is $Deltax=0.45-0.27=0.18m$

The spring constant is

$k=F/(Deltax)=(4g/0.18)Nm^-1$

$=217.8kgs^-2$

as

$1 Nm^-1=1 kgms^-2/m=1 kgs^-2$

If the length of a 27 cm27 cm spring increases to 45 cm45 cm when a 4 kg4 kg weight is hanging from it, what is the spring's constant?