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 determine the potential energy stored in a spring when stretched, we use the formula for potential energy of a spring: \( U = \frac{1}{2} k x^2 \), where \( U \) is the potential energy, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.

Initially, the spring is stretched by 2 cm, which is the displacement \( x_1 \), and the potential energy is \( U \). Using the formula: \( U = \frac{1}{2} k (2)^2 = 2k \).

Now, consider the spring stretched by 8 cm, which is the displacement \( x_2 \). The potential energy stored is given by: \( U_2 = \frac{1}{2} k (8)^2 = 32k \).

To find the new potential energy in terms of \( U \), we set up the equation from earlier: \( U = 2k \), so \( k = \frac{U}{2} \). Substitute this back into the expression for \( U_2 \):

\[ U_2 = \frac{1}{2} \frac{U}{2} \cdot 64 = 16U \]

Thus, if the spring is stretched by 8 cm, the potential energy stored in it is \( 16U \).

Answer 2

The correct option is (A): 16U

The potential energy of a spring = \(\frac{1}{2}x\)\(\times\)force constant\(\times\)(extension)²

potential energy \(\propto\) (extension)²

or, \(\frac{U_1}{U_2}=(\frac{x_1}{x_2})^2=(\frac{2}{8})^2=\frac{1}{16}\)

or, \(U_2=16\,U_1=16U\)(\(\because U_1=U\))