A student stretches a spring horizontally a distance of 20 mm by applying a force of 0.20 N [E]. Determine the force constant of the spring?

1 Answer
Truong-Son N.
Apr 1, 2017

This is going to involve Hooke's law:

#vecF = -kDeltavecx#,

where:

  • #Deltavecx# is the displacement of the spring from its equilibrium position. We consider this to be positive.
  • #k# is the force constant in either #"N/m"# or #"kg/s"^2#. In this case it is more useful in #"N/m"#. It is also necessarily nonnegative (#>=0#).
  • #vecF# is known as the restoring force, and counteracts the natural motion of the spring. Thus, we place a negative sign in front of #kDeltavecx# to account for that, and treat the force as numerically negative.

A distance of #"20 mm"# is converted to #"m"# as follows:

#"20 mm" xx "1 m"/"1000 mm"#

#=# #"0.020 m"#

So, the force constant is:

#color(blue)(k) = (vecF)/(-Deltavecx)#

#= (-"0.20 N")/(-("0.020 m"))#

#=# #color(blue)("10 N/m")#

The larger #k# is, the stiffer the spring. In this case, the spring is very loose. A typical value of #k# for diatomic molecules is on the order of a few hundred (for single bonds) to a few thousand (for double/triple bonds).

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