Question

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?

Answer 1

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).