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When a $200\,N$ force is applied on an object, its length is increased by $1\,mm$.What is the potential energy stored in the object due to this force?
A. $0.2\,J$
B. $10\,J$
C. $20\,J$
D. $0.1\,J$

Last updated date: 19th Jul 2024
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Hint:Potential energy is the energy in a body stored due to its state of motion. When an object is stretched to some length it can be considered as that energy stored in it as a restoring force and potential energy is given as $P.E = \dfrac{1}{2}k{x^2}$ where $k$ is some proportionality constant with a force as $F = kx$.

Complete step by step answer:
Let us suppose object is applied with a force of $F = 200\,N$ and its length is increased by $x = 1\,mm$
Convert the length in millimetre into metre as: $1mm = {10^{ - 3}}\,m$ so we get,
$F = 200\,N$
$\Rightarrow x = {10^{ - 3}}\,m$
Comparing this with the equation $F = kx$ we can write it’s as:
$k = \dfrac{{200}}{{{{10}^{ - 3}}}}N{m^{ - 1}}$
Now, we know the proportionality constant value,
So, using the formula of potential energy due to increase in length is $P.E = \dfrac{1}{2}k{x^2}$
We have, $P.E = \dfrac{1}{2}k{x^2}$
Put $k = \dfrac{{200}}{{{{10}^{ - 3}}}}N{m^{ - 1}}$ and $x = {10^{ - 3}}m$ in above equation we get,
$P.E = \dfrac{1}{2} \times 200 \times {10^{ - 3}}$
$\therefore P.E = 0.1\,J$
So, the potential energy stored in the object due to increase in its length is $P.E = 0.1\,J$.

Hence, the correct option is D.

Note:Whenever an object length gets increased after applying some force, objects have a tendency to store energy to come back into their original state and this restoring force always acts opposite to that of applied force and hence this restoring energy behaves as potential energy of the body. $F = kx$ This equation is generally referred to as Hooke’s law.