Question

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

• A. U/4
• B. 4U
• C. 8U
• D. 16U

Answer 1

The Correct Option is D

The potential energy stored in a spring is given by the formula:

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

where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

First, let's understand the relationship between the displacement of the spring and the potential energy stored:

1. If the spring is stretched by 2 cm, the potential energy is U:

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

2. Now, if the spring is stretched by 8 cm, the new potential energy, let's call it U_{\text{new}}, is:

U_{\text{new}} = \frac{1}{2} k (8)^2 = 32k \text{ cm}^2

Comparing the two values of potential energy:

U_{\text{new}} = 16U

This indicates that the potential energy stored in the spring when stretched by 8 cm is 16 times the potential energy stored when stretched by 2 cm.

Conclusion: The potential energy stored in the spring when stretched by 8 cm is 16U.