## Question

###### If the length of a #21 cm# spring increases to #57 cm# when a #5 kg# weight is hanging from it, what is the spring's constant?

If the length of a #21 cm# spring increases to #57 cm# when a #5 kg# weight is hanging from it, what is the spring's constant?

## Answers

We should expect#k# to be somewhere in the hundreds for typical springs.This is simply looking at the equation

#\mathbf(F = -kDeltay)# ...for when

an object is hanging off a spring.So,

the only force acting on the spring is the force#\mathbf(F_g)# due to gravity#\mathbf(g)# . The generic force#F# , then, is#F_g# . Thus, with#g < 0# (where down is negative):

#F_g = -kDeltay#

#mg = -k(y_f - y_i)#

#(mg)/(y_i - y_f) = k#

#color(blue)(k) = (("5 kg")(-"9.80665" cancel"m""/s"^2))/("0.21" cancel"m" - "0.57" cancel"m")#

#= color(blue)("136.20 kg/s"^2)# or#color(blue)("N/m")#

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