Question

A system consists of two springs connected in series and each having the spring constant \(10\,\text{N m}^{-1}\). The minimum work required to stretch this system by \(1\,\text{cm}\) in erg is

• A. \(1500\)
• B. \(2000\)
• C. \(3000\)
• D. \(2500\)

Hint:

For springs in series, \[ \frac{1}{k_{\text{eq}}}=\frac{1}{k_1}+\frac{1}{k_2}. \] The work done in stretching a spring is \[ W=\frac{1}{2}kx^2. \]

Answer 1

The Correct Option is D

Step 1: Find equivalent stiffness for series springs.
1/k_eq = 1/10 + 1/10 = 2/10 → k_eq = 5 N/m.

Step 2: Compute elastic potential energy.
W = ½ k_eq x² = ½(5)(0.01)² = 2.5×10⁻⁴ J.

Step 3: Convert to ergs.
2.5×10⁻⁴ × 10⁷ = 2500 erg.

Step 4: Final Answer:
2500 erg.