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A long spring is stretched by $2cm$ and its potential energy is $U$. If the spring is stretched by $10cm$ ; its potential energy will be:
A) $U/5$
B) $U/25$
C) $5U$
D) $25U$

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Last updated date: 09th Apr 2024
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MVSAT 2024
Answer
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Hint: First, calculate the value of the spring constant using the formula for energy in a spring by putting the given values, i.e. $x = 2$ and $PE = U$ . Then use the same formula by putting $x = 10cm$ and spring constant as calculated previously to get the value of final potential energy. This will be our final answer.

Formula Used:
Potential energy in a spring, $PE = \dfrac{1}{2}k{x^2}$
Where, $k$ is the spring constant of the spring and $x$ the distance by which the string has been stretched, or extended.

Complete step by step solution:
First, we will use the formula for energy in a spring and put $x = 2$ , as given to get the value of spring constant $k$ . Once we find the value of $k$ we will move on to find the potential energy of the spring when it is stretched by $10cm$ , which will be the final answer.
We have, potential energy in a spring, $PE = \dfrac{1}{2}k{x^2}$
Where, $k$ is the spring constant of the spring and $x$ the distance by which the string has been stretched.
For $x = 2$ and $PE = U$ (given) we get $U = \dfrac{1}{2}k{(2)^2}$
$ \Rightarrow k = \dfrac{{2U}}{4} = \dfrac{U}{2}$
Now we will use this value of spring constant in the formula of potential energy of the spring with $x = 10cm$ to calculate the potential energy of spring in this case.
Therefore, we get $PE = \dfrac{1}{2} \times \dfrac{U}{2} \times {(10)^2}$
On simplifying, we are left with $PE = \dfrac{U}{4} \times 100$
Which gives, $PE = 25U$
Hence, option D is the correct answer.

Note: In questions like these, make sure both the values of extension of spring are given with the same unit. If the units are different, make sure you convert the given values into the same unit before using the values in an answer. Otherwise, you may end up getting the wrong answer. A spring stores energy as potential energy and releases the stored energy as kinetic energy. The kinetic energy released is proportional to the square of the length by which the spring is compressed or stressed. Thus, the potential energy is also directly proportional to the square of the length by which the spring is compressed or stressed.