Question

When a long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy stored in it will be:

• A. \(\frac {U}{5}\)
• B. \(5U\)
• C. \(10U\)
• D. \(25U\)

Answer 1

The Correct Option is D

To determine the potential energy stored when a spring is stretched, we can use Hooke's law, which states that the potential energy \( E \) stored in a spring is given by the equation:

E = \frac{1}{2} k x^2

where:

• k is the spring constant, and
• x is the amount of stretch (or compression) from the equilibrium position.

We know that when the spring is stretched by 2 cm, the potential energy is \( U \). Thus, for a stretch of 2 cm, we can write:

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

The formula simplifies to:

U = 2k

Next, if the spring is stretched by 10 cm, the potential energy is:

E' = \frac{1}{2} k (10)^2

This simplifies to:

E' = 50k

To find how \( E' \) relates to \( U \), divide the second equation by the first equation:

\frac{E'}{U} = \frac{50k}{2k} = 25

Therefore:

E' = 25U

Thus, the potential energy stored in the spring when it is stretched by 10 cm is 25U.

Correct Answer: 25U