Question:

A long spring, when stretched by a distance $ z $, has potential energy $ U $. On increasing the stretching to $ n $ times, the potential energy of the spring will be

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The potential energy stored in a spring is proportional to the square of the displacement from the equilibrium position. So, increasing the displacement by a factor of $ n $ increases the potential energy by a factor of $ n^2 $.

Updated On: May 3, 2025

$ \frac{U}{n} $
$ nU $
$ n^2 U $
$ \frac{U}{n^2} $

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The Correct Option is C

Approach Solution - 1

To determine how the potential energy of a spring changes when the stretching is increased, we can use Hooke's Law and the formula for the potential energy of the spring. The potential energy $U$ stored in a spring is given by:

$U = \frac{1}{2}kz^2$

where:

$k$ is the spring constant.
$z$ is the amount of stretch or compression from the spring's equilibrium position.

Given that initially the spring is stretched by a distance $z$ and the potential energy is $U$:

$U = \frac{1}{2}kz^2$

If the stretching is increased to $n$ times, then the new stretching distance will be $nz$. The new potential energy $U'$ is given by:

$U' = \frac{1}{2}k(nz)^2$

Which simplifies to:

$U' = \frac{1}{2}k \cdot n^2 \cdot z^2$

Comparing the original and new potential energies:

$U' = \frac{1}{2}k \cdot n^2 \cdot z^2 = n^2 \cdot \frac{1}{2}kz^2 = n^2 \cdot U$

Thus, the potential energy of the spring when stretched to $n$ times the original distance is $n^2U$.

Therefore, the correct answer is: $ n^2U $

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Approach Solution -2

The potential energy $ U $ stored in a spring is given by Hooke's law for potential energy: \[ U = \frac{1}{2} k z^2 \] where:
- $ k $ is the spring constant,
- $ z $ is the displacement from the equilibrium position. If the spring is stretched by a factor of $ n $ (i.e., the new displacement is $ n \times z $), the new potential energy $ U' $ will be: \[ U' = \frac{1}{2} k (nz)^2 = n^2 \cdot \frac{1}{2} k z^2 = n^2 U \]
Thus, the potential energy increases by a factor of $ n^2 $. Therefore, the correct answer is: \[ \text{(3) } n^2 U \]