Question

A spring stores 1 J of energy for a compression of 1 mm. The additional work to be done to compress it further by 1 mm is

• A. \(1 \text{ J}\)
• B. \(2 \text{ J}\)
• C. \(3 \text{ J}\)
• D. \(4 \text{ J}\)
• E. \(0.5 \text{ J}\)

Hint:

For equal successive increments of compression, the work done in each interval follows the ratio of odd numbers: \(1:3:5:7...\).

Answer 1

The Correct Option is C

Concept: The potential energy stored in a spring is given by \(U = \frac{1}{2}kx^2\).
• Initial work/energy: \(W_1 = \frac{1}{2}kx_1^2\)
• Total work for total compression: \(W_{total} = \frac{1}{2}kx_{final}^2\)
• Additional work: \(\Delta W = W_{total} - W_1\)

Step 1: Find the total energy for the final compression.
Initially, \(x_1 = 1 \text{ mm}\) and \(U_1 = 1 \text{ J}\). Final compression \(x_2 = 1 \text{ mm} + 1 \text{ mm} = 2 \text{ mm}\). Since \(U \propto x^2\): \[ \frac{U_2}{U_1} = \left(\frac{x_2}{x_1}\right)^2 = \left(\frac{2}{1}\right)^2 = 4 \implies U_2 = 4 \text{ J} \]

Step 2: Calculate the additional work.
\[ \text{Additional Work} = U_2 - U_1 = 4 \text{ J} - 1 \text{ J} = 3 \text{ J} \]