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

• A. \( \frac{U}{n} \)
• B. \( nU \)
• C. \( n^2 U \)
• D. \( \frac{U}{n^2} \)

Hint:

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 \( n^2 \).

Answer 1

The Correct Option is C

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: \( n^2U \)

Answer 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: \[ \text{(3) } n^2 U \]