Question

A 540 gram object is attached to a vertical spring, causing the spring’s length to change from 70 cm to 110 cm.

What is the spring constant k?

Group of answer choices

130 N/m

33 N/m

13 N/m

330 N/m

Answer 1

Approximately
`13\; {\rm N \cdot m^(-1)}` (assuming that
`g = 9.81\; {\rm N \cdot kg^(-1)}`.)

Step-by-step explanation:

Let
`F_{\text{s}}` denote the force that this spring exerts on the object. Let
`x` denote the displacement of this spring from the equilibrium position.

By Hooke's Law, the spring constant
`k` of this spring would ensure that
`F_\text{s} = -k\, x`.

Note that the mass of the object attached to this spring is
`m = 540\; {\rm g} = 0.540\; {\rm kg}`. Thus, the weight of this object would be
`m\, g = 0.540\; {\rm kg} * 9.81\; {\rm N \cdot kg^(-1)} \approx 5.230\; {\rm N}`.

Assuming that this object is not moving, the spring would need to exert an upward force of the same magnitude on the object. Thus,
`F_{\text{s}} = 5.230\; {\rm N}`.

The spring in this question was stretched downward from its equilibrium by:

`\begin{aligned} x &= (70\; {\rm cm} - 110\; {\rm cm}) \\ &= (-40)\; {\rm cm} \\ &= (-0.40) \; {\rm m}\end{aligned}`.

(Note that
`x` is negative since this displacement points downwards.)

Rearrange Hooke's Law to find
`k` in terms of
`F_{\text{s}}` and
`x`:

`\begin{aligned} k &= \frac{F_{\text{s}}}{-x} \\ &\approx \frac{5.230\; {\rm N}}{-(-0.40)\; {\rm m}} \\ &\approx 13\; {\rm N \cdot m^(-1)}\end{aligned}`.