If the length of a $20 cm$ spring increases to $56 cm$ when a $15 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2017-07-12T06:00:09

K= 408 N/m

$F= mg = 15 kg xx 9,8 m/(s^2)= 147 N$
For a spring $F= K (X_2-X_1)$ so $K= (F/(X_2-X_1))=(147 N)/((56-20) cm)= (147N)/(0,36 m)= 408 N/m$

If the length of a 20 cm20 cm spring increases to 56 cm56 cm when a 15 kg15 kg weight is hanging from it, what is the spring's constant?