Question

The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be:

• A. 16U
• B. 2U
• C. 4U
• D. 8U

Answer 1

The Correct Option is A

To solve the problem of determining the potential energy stored in a spring when it is stretched, we can use the formula for the potential energy stored in a spring, which is given by:

\(U = \frac{1}{2} k x^2\)

Where:

• \(U\) is the potential energy.
• \(k\) is the spring constant.
• \(x\) is the displacement from the equilibrium position (the amount the spring is stretched or compressed).

Given that the potential energy of the spring when stretched by 2 cm (0.02 m) is \(U\), we can write:

\(U = \frac{1}{2} k (0.02)^2\)

Now, we need to find the potential energy when the spring is stretched by 8 cm (0.08 m). Let's denote this potential energy as \(U'\):

\(U' = \frac{1}{2} k (0.08)^2\)

To compare \(U'\) to \(U\), we calculate the ratio:

\(\frac{U'}{U} = \frac{\frac{1}{2} k (0.08)^2}{\frac{1}{2} k (0.02)^2}\)

Notice that the spring constant \(k\) and the factor \(\frac{1}{2}\) cancel out in the numerator and denominator, simplifying to:

\(\frac{U'}{U} = \frac{(0.08)^2}{(0.02)^2} = \frac{0.0064}{0.0004} = 16\)

This means that the potential energy \(U'\) when the spring is stretched by 8 cm is 16 times that of \(U\) when stretched by 2 cm.

Thus, the potential energy stored in the spring when stretched by 8 cm is 16U.

Therefore, the correct answer is 16U.