To calculate the work done in stretching the wire by 1 mm using a constant external force, we can follow these steps:

Given dataTo calculate the work done in stretching the wire by 1 mm using a constant external force, we can follow these steps:

Given

• Cross-sectional area, $A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2$
• Initial length, $L = 1 \, \text{m}$
• Extension, $\Delta L = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m}$
• Young's modulus, $Y = 2 \times 10^{11} \, \text{N/m}^2$

Step 1: Relation between force and extension (Hooke's Law)

From Hooke's law, the force $F$ applied to stretch the wire is given by: $F = \frac{Y A \Delta L}{L}$ Substitute the values: $F = \frac{(2 \times 10^{11}) (1 \times 10^{-4}) (1 \times 10^{-3})}{1}$ $F = 2 \times 10^4 \, \text{N}$

Step 2: Work done in stretching the wire

The work done $W$ to stretch the wire by an amount $\Delta L$ is given by the formula: $W = \frac{1}{2} F \Delta L$ Substitute the values of $F$ and $\Delta L$: $W = \frac{1}{2} (2 \times 10^4) (1 \times 10^{-3})$ $W = 10 \, \text{J}$

Final Answer:

The amount of work done by the constant external force in increasing the length of the wire by 1 mm is $\boxed{10 \, \text{J}}$.

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Would you like more details or have any further questions?

Here are some related questions for practice:

1. How does the Young's modulus of a material affect the work done during stretching?
2. What would be the work done if the extension was doubled to 2 mm?
3. How does the cross-sectional area impact the force required for a given extension?
4. What is the significance of Young's modulus in material science?
5. If the wire was made of a different metal with half the Young's modulus, how would the work change?

Tip: Young's modulus provides a measure of how stiff a material is—higher values mean the material is less likely to deform under stress.