Question

A spring with force constant \(k\) is initially stretched by \(x_1\). If it is further stretched by \(x_2\), then the increase in its potential energy is

• A. \(\frac{1}{2}k(x_2 - x_1)^2\)
• B. \(\frac{1}{2}k x_2 (x_2 + 2x_1)\)
• C. \(\frac{1}{2}k x_1^2 - \frac{1}{2}k x_2^2\)
• D. \(\frac{1}{2}k (x_1 + x_2)^2\)
• E. \(\frac{1}{2}k (x_1^2 + x_2^2)\)

Hint:

Always calculate change in spring energy using \(U_f - U_i\), not directly using final formula.

Answer 1

The Correct Option is B

Concept: The potential energy stored in a spring is given by: \[ U = \frac{1}{2}kx^2 \] The increase in potential energy is the difference between final and initial energies.

Step 1: Initial extension. \[ x = x_1 \Rightarrow U_1 = \frac{1}{2}k x_1^2 \]

Step 2: Final extension. Further stretched by \(x_2\), so total extension: \[ x = x_1 + x_2 \Rightarrow U_2 = \frac{1}{2}k(x_1 + x_2)^2 \]

Step 3: Increase in potential energy. \[ \Delta U = U_2 - U_1 = \frac{1}{2}k[(x_1 + x_2)^2 - x_1^2] \]

Step 4: Expand expression. \[ (x_1 + x_2)^2 = x_1^2 + 2x_1x_2 + x_2^2 \] \[ \Delta U = \frac{1}{2}k(x_2^2 + 2x_1x_2) \]

Step 5: Final simplified form. \[ \Delta U = \frac{1}{2}k x_2 (x_2 + 2x_1) \]

Step 6: Conclusion. \[ \boxed{\frac{1}{2}k x_2 (x_2 + 2x_1)} \]