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: Recall formula for work done in stretching a spring.
Work done by a spring:
\[ W = \frac{1}{2} k (x_2^2 - x_1^2) \]
where \(x_1, x_2\) are extensions from natural length.

Step 2: Compute extensions.
Natural length \(l_0 = 0.25 \, \text{m}\), initial length \(x_1 = 0.50 - 0.25 = 0.25 \, \text{m}\), final length \(x_2 = 0.60 - 0.25 = 0.35 \, \text{m}\).

Step 3: Substitute values.
\[ W = \frac{1}{2} \cdot 50 \cdot (0.35^2 - 0.25^2) \]

Step 4: Simplify the squares.
\[ 0.35^2 = 0.1225, \quad 0.25^2 = 0.0625 \]
\[ 0.1225 - 0.0625 = 0.06 \]

Step 5: Calculate work.
\[ W = \frac{1}{2} \cdot 50 \cdot 0.06 = 25 \cdot 0.06 = 1.5 \, \text{J} \]

Step 6: Final conclusion.
Hence, the work done in stretching the spring is:
\[ \boxed{1.5 \, \text{J}} \]