Question

A vertical spring has a length of 0.225 m when a 0.25 kg mass hangs from it, and a length of 0.75 m when a 1.975 kg mass hangs from it.

a) Find the spring constant k in N/m.

b) Find the unloaded length of the spring in cm

Answer 1

The spring constant is approximately
`32\; {\rm N\cdot m^(-1)}` (assuming that
`g = 9.81\; {\rm N\cdot kg^(-1)}`.)

The unloaded length of the spring is approximately
`15\; {\rm cm}`. (Approximately
`0.15\; {\rm m}`.)

Step-by-step explanation:

By Hooke's Law, the restoring force from an ideal spring is equal to the product of the spring constant
`k` and the displacement
`x` from the unloaded position of the spring.

Let
`k` denote the spring constant (in
`\rm N\cdot m^(-1)}`) of this spring and let
`L` denote the unloaded length in meters.

When the spring is stretched to a length of
`0.225\; {\rm m}` from the unloaded length of
`L`, the displacement of this spring from the unloaded position would be
`(0.225 - L)` (meters.) By Hooke's Law, the spring would exert a restoring force of
`k\, (0.225 - L)` on the
`m = 0.25\; {\rm kg}` mass. This force should be equal in magnitude to the weight of the mass,
`m\, g = (0.25)\, (9.81)\; {\rm N}`:

`k\, (0.225 - L) = (0.25)\, (9.81)`.

Similarly, when the
`m = 1.975\; {\rm kg}` mass is attached to the spring, the displacement of the spring from the unloaded position would be
`(0.75 - L)`, and the restoring force would be
`k\, (0.75 - L)`. The weight of this mass would be
`(1.975)\, (9.81)\; {\rm N}`. Thus:

`k\, (0.75 - L) = (1.975)\, (9.81)`.

Solve this system of equations for
`k` and
`L`:

`\left\lbrace\begin{aligned}& k\, (0.225 - L) = (0.25)\, (9.81) \\ &k\, (0.75 - L) = (1.975)\, (9.81)\end{aligned}\right.`.

`\left\lbrace\begin{aligned}& k \approx 32 \\ & L \approx 0.15\end{aligned}\right.`.

In other words, the spring constant is approximately
`32\; {\rm N\cdot m^(-1)}`. The unloaded length of the spring is approximately
`0.15\; {\rm m}`, which is equivalent to approximately
`15\; {\rm cm}`.