If the length of a #45 cm# spring increases to #63 cm# when a #7 kg# weight is hanging from it, what is the spring's constant?

1 Answer
Jun 11, 2018

Answer:

Approximately #381.1# newtons per meter.

Explanation:

We use Hooke's law, which states that,

#F=kx#

where:

  • #F# is the force applied in newtons

  • #k# is the spring constant in #"kg/s"^2# or #"N/m"#

  • #x# is the extension in meters

Here, the force is the weight of the object, which is equal to:

#7 \ "kg"*9.8 \ "m/s"^2=68.6 \ "N"#.

The spring constant is then:

#k=F/x#

#=(68.6 \ "N")/(0.63 \ "m"-0.45 \ "m")#

#=(68.6 \ "N")/(0.18 \ "m")#

#~~381.1 \ "N/m"#