Final answer:
To find the change in length of the pencil lead when a 4.0 N force is applied, we calculate area, stress, strain, and then change in length, but the answer of 1.222236 mm decrease seems unreasonable given normal use of pencils.
Step-by-step explanation:
To calculate how much shorter the length of the pencil lead becomes, we apply Hooke's Law and the definition of Young's modulus (E). The Young's modulus is defined as the ratio of stress to strain, where stress is force per unit area, and strain is the deformation (change in length) over original length.
Let's calculate the area for the circular cross-section of the pencil lead first:
- Area (A) = π * (diameter/2)^2 = 3.14159 * (0.50 mm / 2)^2 = 0.19635 mm², which is 0.19635 * 10^-6 m² in SI units.
Now, we use the formula Stress = Force / Area to find stress:
- Stress (σ) = Force (F) / Area (A) = 4.0 N / 0.19635 * 10^-6 m² = 2.03706 * 10^7 N/m².
Young's modulus (E) is related to stress and strain (ε) by E = σ / ε, so:
- Strain (ε) = Stress (σ) / Young's modulus (E) = 2.03706 * 10^7 N/m² / 1 * 10^9 N/m² = 0.0203706.
Finally, we calculate the change in length (ΔL) using the formula ΔL = Strain (ε) * Original length (L):
- Change in length (ΔL) = Strain (ε) * Original Length (L) = 0.0203706 * 60 mm = 1.222236 mm.
As for the reasonability of the result, it would be unusual to observe a pencil lead shorten by over 1 mm under a 4 N force in practice. Typically, pencil leads don't exhibit that level of elastic deformation under normal writing pressure, so the calculated change in length seems excessive, suggesting an error in the given modulus value or the need for a more detailed model that accounts for other factors like the brittle nature of graphite.