Question:
A wire is stretched by \( 2 \) mm under force \( 100 \) N. Elastic energy stored is:
A wire is stretched by \( 2 \) mm under force \( 100 \) N. Elastic energy stored is:
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The factor of \( \frac{1}{2} \) is crucial because the force increases linearly from \( 0 \) to \( F \) as the wire stretches. Think of it as using the average force (\( F/2 \)) over the total distance of the stretch.
Updated On: Jun 3, 2026
- \( 0.05 \) J
- \( 0.1 \) J
- \( 0.2 \) J
- \( 0.01 \) J
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The Correct Option is B
Solution and Explanation
Concept:
Elastic potential energy is the energy stored as a result of applying a force to deform an elastic object. For a stretched wire, the work done by the stretching force is stored as elastic potential energy (\( U \)). If the deformation is within the elastic limit, the formula is:
\[ U = \frac{1}{2} \times \text{Force} \times \text{Extension} \]
This formula represents the area under the Force-Extension graph.
Step 1: Extracting given values from and converting to SI units.
Force (\( F \)) = \( 100 \) N
Extension (\( \Delta L \)) = \( 2 \) mm = \( 2 \times 10^{-3} \) m
Step 2: Substituting values into the elastic energy formula.
\[ U = \frac{1}{2} \times F \times \Delta L \] \[ U = \frac{1}{2} \times 100 \times (2 \times 10^{-3}) \]
Step 3: Calculating the final result.
\[ U = 100 \times 10^{-3} \] \[ U = 0.1 \, J \]
Step 1: Extracting given values from and converting to SI units.
Force (\( F \)) = \( 100 \) N
Extension (\( \Delta L \)) = \( 2 \) mm = \( 2 \times 10^{-3} \) m
Step 2: Substituting values into the elastic energy formula.
\[ U = \frac{1}{2} \times F \times \Delta L \] \[ U = \frac{1}{2} \times 100 \times (2 \times 10^{-3}) \]
Step 3: Calculating the final result.
\[ U = 100 \times 10^{-3} \] \[ U = 0.1 \, J \]
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