Question

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

Answer 1

Answer:

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

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 $m\, g = 0.540\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \approx 5.230\; {\rm N}$.

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 $x$:

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