Question

A 10-cm-long spring is attached to theceiling. When a2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.

a.What is the spring constant k?

b.How long is the spring when a 3.0kg mass is suspended from it?

Answer 1

Concepts and reason

Concepts used here is the expression of force, Hooke’s law and equilibrium.

The object is hanging to the ceiling through spring. A spring stretches to a certain length. Initial and final length of the spring are given. At equilibrium, the restoring force is equal to weight of the body, then the value of spring constant can be calculated.

Fundamentals

In case of elasticity, equilibrium happens when restoring force is equal to weight of the body.

$mg = k\Delta x$

Here, $m$ is the mass of the body, $g$ is the acceleration due to gravity, $k$ is the spring constant, and $\Delta x$ is the extension or change in length.

The change in length is given as,

$\Delta x = {l_{\rm{f}}} - {l_{\rm{i}}}$

Here, ${l_{\rm{f}}}$ is the final length and ${l_{\rm{i}}}$ is the initial length.

(a)

The change in the length is calculated as,

$\Delta x = {l_{\rm{f}}} - {l_{\rm{i}}}$

Substitute $10 {\rm{cm}}$ for ${l_{\rm{i}}}$ and $15 {\rm{cm}}$ for ${l_{\rm{f}}}$ in the expression of $\Delta x$ .

$\begin{array}{c}\\\Delta l = \left( {15 {\rm{cm}}} \right) - \left( {10 {\rm{cm}}} \right)\\\\ = \left( {5 {\rm{cm}}} \right)\left( {\frac{{{\rm{1}} {\rm{m}}}}{{{\rm{100}} {\rm{cm}}}}} \right)\\\\ = 0.05 {\rm{m}}\\\end{array}$

At equilibrium, the net force acting on the object is zero. Therefore, balance the forces acting on the object in the vertical direction.

$mg = k\Delta x$

Rearrange the equation $mg = k\Delta x$ for $k$ ,

$k = \frac{{mg}}{{\Delta x}}$

Substitute 2 kg for $m$ , ${\rm{9}}{\rm{.8}} {\rm{m}} \cdot {{\rm{s}}^{ - 2}}$ for $g$ , and 0.05 m for $\Delta x$ in the expression $k = \frac{{mg}}{{\Delta x}}$ .

$\begin{array}{c}\\k = \frac{{\left( {{\rm{2}} {\rm{kg}}} \right)\left( {{\rm{9}}{\rm{.8}} {\rm{m}} \cdot {{\rm{s}}^{ - 2}}} \right)}}{{{\rm{0}}{\rm{.05}} {\rm{m}}}}\\\\ = 392 {\rm N} \cdot {{\rm{m}}^{ - 1}}\\\end{array}$

(b)

Use equilibrium equation to absolve the new value of $\Delta x$ .

$mg = k\Delta x$

Rearrange the equation $mg = k\Delta x$ for $\Delta x$ ,

$\Delta x = \frac{{mg}}{k}$

Substitute ${\rm{3}} {\rm{kg}}$ for $m$ , ${\rm{9}}{\rm{.8}} {\rm{m}} \cdot {{\rm{s}}^{ - 2}}$ for $g$ , and ${\rm{392}} {\rm{N}} \cdot {{\rm{m}}^{ - 1}}$ for $k$ in the expression $\Delta x = \frac{{mg}}{k}$ .

$\begin{array}{c}\\\Delta x = \frac{{\left( {{\rm{3}} {\rm{kg}}} \right)\left( {{\rm{9}}{\rm{.8}} {\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)}}{{{\rm{392}} {\rm{N}} \cdot {{\rm{m}}^{ - 1}}}}\\\\ = {\rm{0}}{\rm{.075}} {\rm{m}}\\\end{array}$

The new final length of the spring can be calculated as,

$\begin{array}{c}\\\Delta x = {l_f} - {l_i}\\\\{l_f} = {l_i} + \Delta x\\\end{array}$

Substitute ${\rm{0}}{\rm{.1}} {\rm{m}}$ for ${l_i}$ and ${\rm{0}}{\rm{.075}} {\rm{m}}$ for $\Delta x$ in the change in the length expression.

$\begin{array}{c}\\{l_f} = {l_i} + \Delta x\\\\ = \left( {{\rm{0}}{\rm{.1}} {\rm{m + 0}}{\rm{.075}} {\rm{m}}} \right)\\\\ = 0.175 {\rm{m}}\\\end{array}$

Final length of the spring is ${\rm{0}}{\rm{.175}}\;{\rm{m}}$ .

Ans: Part a

The spring constant is $392 {\rm N} \cdot {{\rm{m}}^{ - 1}}$ .

Part b

Final length of spring is ${\rm{0}}{\rm{.175}}\;{\rm{m}}$ .

Answer 2

Part a

The spring constant is $392 {\rm N} \cdot {{\rm{m}}^{ - 1}}$ .

Part b

Final length of spring is ${\rm{0}}{\rm{.175}}\;{\rm{m}}$ .