Question

The work done in stretching a spring of natural length 25 cm and spring constant 50 N/m from 50 cm to 60 cm is:

• A. 1.5 J
• B. 2 J
• C. 3.5 J
• D. 5 J

Hint:

For a spring, always measure extension from natural length: \(W = \frac{1}{2} k (x_2^2 - x_1^2)\).

Answer 1

The Correct Option is A

Step 1: Use the average force over the extra stretch instead of the energy-difference formula.
Since the spring force grows linearly with extension, the work done while stretching from one extension to another equals the average of the two spring forces times the additional distance stretched.

Step 2: Find the two extensions from the natural length of 25 cm.
\[ x_1 = 0.50 - 0.25 = 0.25\ \text{m}, \qquad x_2 = 0.60 - 0.25 = 0.35\ \text{m} \]

Step 3: Find the spring force at each extension.
\[ F_1 = kx_1 = 50\times0.25 = 12.5\ \text{N}, \qquad F_2 = kx_2 = 50\times0.35 = 17.5\ \text{N} \]

Step 4: Multiply the average force by the extra stretch, $x_2 - x_1 = 0.10$ m.
\[ W = \frac{F_1+F_2}{2}\times(x_2-x_1) = \frac{12.5+17.5}{2}\times0.10 = 15\times0.10 \]

Step 5: Conclusion.
\[ W = \boxed{1.5\ \text{J}} \]