Question

A spring with a spring constant of $144N{m^{ - 1}}$ is compressed by a distance of $16.5cm$ .How much elastic potential energy is stored in the spring?

Answer 1

In order to solve this question, we should know that when a spring is stretched to some distance it stores potential energy to bring it back to its original position, here we will use the general formula of potential energy of spring to solve this problem.

Formula used: If k denoted for the spring constant of a spring and x is the distance to which spring is stretched from its original position, then stored potential energy is calculated using formula
$P.E = \dfrac{1}{2}k{x^2}$ where P.E stands for potential energy.

Complete step by step answer:
According to the question, we have given that
$k = 144N{m^{ - 1}}$ spring constant of the spring
$x = 16.5cm$ distance to which spring is stretched from original position.
Here, distance is given in centimetres so, let us first convert the distance from centimetre to metres as
$x = 0.165m$
now, using the formula of potential energy of the spring, we have
$P.E = \dfrac{1}{2}k{x^2}$ on putting the value of parameter K and parameter x we get,
$P.E = \dfrac{1}{2}(144){(0.165)^2}$
on further solving we get,
$P.E = 72 \times 0.027$
$P.E = 1.94Nm$
and Newton-meter is a unit of energy which can also be written as Joules.
Hence, the potential energy stored in the spring is $P.E = 1.94Joules$.

Note: It should be remembered that, while solving such questions always convert all the parameters value in same units and here unit of conversion is used as $1cm = 0.01m$ and potential energy of a spring is also sometimes known as restoring energy which shows that it’s the energy which spring use to restore its original position.