Question

A wire 4m long 0.3mm diameter is stretched by a force of 8 kg wt. If the extension in the length amounts to 1.5mm, calculate the energy stored in the wire?

Answer 1

Hint: Energy is defined as the ability or capacity of any object or body to do some work. In this case, as the wire is stretched and hence there is increase in the length of the wire, therefore, the energy stored in this case will be in the form of potential energy.

Complete step by step solution:
Given that the length of the wire is 4m.
Force = 8kg wt
Or it can be written as \[8 \times 9.8 = 78.4N\].
After the wire is stretched, the extension in the length is \[\Delta l = 1.5mm\].
Or it can be written as \[\Delta l = 1.5 \times {10^{ - 3}}m\].

When an object such as a wire or a spring is stretched or compressed, a form of energy also known as elastic potential energy is stored in that object. The elastic potential energy stored in the wire due to stretching can be calculated using the following formula,
\[E = \dfrac{1}{2} \times F \times \Delta l\]
Substituting all the given values in the above formula and solving for the value of energy, we get
\[E = \dfrac{1}{2} \times 78.4 \times 1.5 \times {10^{ - 3}}\]
\[\therefore E = 58.8 \times {10^{ - 3}}J\]

Therefore, the elastic potential energy stored in the wire due to it being stretched will be equal to \[58.8 \times {10^{ - 3}}J\].

Note: It is important to remember that there are different forms of energy such as kinetic energy, potential energy, mechanical energy etc. The type of energy and the formula to be used, depends on the situation given, as in this case the wire is stretched and hence the concept of elastic potential energy is used. In case of elastic potential energy, the amount of energy stored depends on the extent to which the object is stretched or compressed.