Question

If a spring has a constant of $2kg/{s^2}$, how much work will it take to extend the spring by $88cm$?

Answer 1

The given problem can be solved with the help of a formula that gives the relationship between stress and strain. Remember that the restoring force is always proportional to the magnitude of the deformation.
Formula used:
$\Rightarrow W = \dfrac{1}{2}k(x_f^2 - x_i^2)$
Where,
$W$is the work done
$x$ is the magnitude of the displacement from the equilibrium exhibited when the string is stretched.
${x_i}$ is the initial displacement
${x_f}$ is the final displacement
$k$ is the proportionality constant which is also known as the spring constant.

Complete step by step answer:
The question is based on the concept of Hooke's law. Consider the material. If the material behaves elastically and it exhibits a linear relationship between the strain and stress, then the material is said to be linearly elastic. When this property shows up then the stress will be directly proportional to the strain.
It can also be said as for smaller deformation stress is directly proportional to strain. In simple words, the hooke's law states that the strain in the solid is directly proportional to the stress that is applied within the elastic limit of that solid. Mathematically it can be expressed as,
$\Rightarrow F = - k.x$
Where,
$F$ is the restoring force
$x$is the magnitude of the displacement from the equilibrium exhibited when the string is stretched.
$k$is the proportionality constant which is also known as the spring constant.
Now let us try to solve the given problem. The value spring constant $k$ is $2kg/{s^2}$. The displacement value $x$ is $88cm$.
The formula is,
$\Rightarrow W = \dfrac{1}{2}k(x_f^2 - x_i^2)$
We need to calculate the work done. The final displacement value is $88cm$ and the initial displacement value is zero.
Substitute the given values in the formula,
$\Rightarrow W = \dfrac{1}{2} \times 2\left[ {{{\left( {88cm} \right)}^2} - 0} \right]$
Convert the centimeter into meter we get,
$\Rightarrow W = \dfrac{1}{2} \times 2\left[ {{{\left( {\dfrac{{88}}{{100}}} \right)}^2} - 0} \right]$
Cancel out the common terms.
$\Rightarrow W = 1 \times \left[ {{{\left( {\dfrac{{88}}{{100}}} \right)}^2} - 0} \right]$
Divide the terms inside the bracket, we get
$\Rightarrow W = 1 \times \left[ {{{\left( {0.88} \right)}^2} - 0} \right]$
Squaring the terms inside the bracket,
$\Rightarrow W = 1 \times \left[ {0.77 - 0} \right]$
By subtracting we get,
$\Rightarrow W = 1 \times 0.77$
Multiply to get the answer.
$\Rightarrow W = 0.77J$
$\therefore W = 0.77J$
Therefore the $0.77J$ work will take to extend the spring.

Note:
Hooke's law is used in all branches of science and engineering. The law is not applicable when the material’s elastic limit exceeds. It is only accurate for the solid bodies when the force and the deformations are small. It is the fundamental principle for the manometer, spring scale, etc.