Question

A spring when compressed by $4{\text{ }}cm$ has $2J$ energy stored in it. The force required to extend it by $8{\text{ }}cm$ will be ________.

Answer 1

A material with the ability to expand or compress and then revert to its original form is said to be a spring. A force may be a pull or a push on any entity caused by its contact with another entity. In the case given in the question the spring (first entity) experiences the force applied by a person (second entity).

Formulas used:
$|{F_s}| = k \times x$
$|{U_s}| = \dfrac{1}{2} \times k \times {x^2}$

Complete step by step answer:
We first understand what is given to us; there is a length to which the string was compressed that is ${x_1} = 4\;cm = 4 \times {10^{ - 2}}\;m$ (in the standard unit). There is a stored value of energy in the string when compressed, it is ${U_s} = 2J$. Another spring length when stretched is given as ${x_2} = 8\;cm = 8 \times {10^{ - 2}}m$ (in the standard unit). Our aim is to find the amount of force needed to stretch this spring to a certain distance, so force $F = ?$

In brief what we need to do is that, since an energy stored in the spring (when it is compressed to a particular length) is given, this is termed to be the given spring’s potential energy. Then we are required to find a specific value of force we need to exert upon that same spring, to stretch it by another given length. This force can be computed using the Hooke’s law where we just require the length to stretch it and the spring constant.

So firstly, let us find the spring constant of the given spring using the spring’s potential energy equation: The elastic potential energy is proportional to the change in length whole squared and the spring's constant.
${U_s} = \dfrac{1}{2} \times k \times {x_1}^2$; here ${U_s} =$ potential energy, ${x^2} =$ square of given length change, $k =$ spring’s constant
Substituting the given values in the potential energy’s equation:
$\Rightarrow 2J = \dfrac{1}{2} \times k \times {4^2} \times {10^{ - 4}}{m^2}$
$\Rightarrow 2J = \dfrac{1}{2} \times k \times 16 \times {10^{ - 4}}{m^2}$

Simplifying by putting the unknown value of ‘spring constant’ on one side:
$\Rightarrow \dfrac{{2J \times 2}}{{16 \times {{10}^{ - 4}}\;{m^2}}} = k$
$\Rightarrow k = 250\;J/{m^2}$
So now our spring constant for the given spring is found to be $250$. Next we require a force and the amount of force used to adjust a spring-like material's length is proportional to the spring's displacement and its spring constant, this is also the Hooke’s law.
$\Rightarrow |{F_s}| = k \times {x_2}$
$\Rightarrow |{F_s}| = 250 \times 8 \times {10^{ - 2}}\;N$
$\therefore |{F_s}| = 200\;N$

Therefore the value of force we need to apply in order to stretch the given spring to $8\;cm$ will be $200\,N$.

Note: When we speak about the potential energy of an object, it is originally the ‘work’ done to that object. For the case of springs, if we draw the graph for spring against its displacement, we can obtain the amount of work spent on the given spring.