Question

A spring has a spring constant 200 Nm\(^{-1}\). If it is stretched by 1 cm then the potential energy stored in it is:

• A. \( 100 \, \text{J} \)
• B. \( 0.01 \, \text{J} \)
• C. \( 10 \, \text{J} \)
• D. \( 1 \, \text{J} \)

Hint:

Ensure consistent SI units (meters for length, Newtons for force) when using spring formulas.

Answer 1

The Correct Option is B

Step 1: Understand potential energy in a spring.
Potential energy is stored when a spring is displaced from its equilibrium position.
Step 2: Identify given values and convert units.
Spring constant \( k = 200 \, \text{Nm}^{-1} \) Extension \( x = 1 \, \text{cm} = 0.01 \, \text{m} \)
Step 3: Apply the potential energy formula.
\( U = \frac{1}{2} k x^2 \)
Step 4: Substitute and calculate.
\( U = \frac{1}{2} (200) (0.01)^2 = \frac{1}{2} (200) (0.0001) = 0.01 \, \text{J} \)