Question

The "lead" in pencils is a graphite composition with a Young’s modulus of about 1 times 10^9 , {N/m}^2. Calculate the change in length of the lead in an automatic pencil if you tap it straight into the pencil with a force of 4.0 N. The lead is 0.50 , {mm} in diameter and 60 , {mm} long. (b) Is the answer reasonable? That is, does it seem to be consistent with what you have observed when using pencils?

A. (a) 0.24 , {mm}, (b) Yes
B. (a) 0.36 , {mm}, (b) No
C. (a) 0.48 , {mm}, (b) Yes
D. (a) 0.60 , {mm}, (b) No

Answer 1

By applying Hooke's Law and the formula for Young's modulus, the change in length of the pencil lead can be calculated using the provided values of force, original length, and cross-sectional area.

erefore, the correct answer is: A. (a) 0.24 mm, (b) Yes

Step-by-step explanation:

To calculate the change in length (ΔL) of the pencil lead when a force (F) is applied, we can use Hooke's Law along with the definition of Young's modulus.

Young's modulus (E) is defined as the stress over strain, where stress is the force applied divided by the cross-sectional area (A), and strain is the change in length (ΔL) divided by the original length (L).

First, we find the cross-sectional area of the pencil lead:

A = πr2 = π(d/2)2

Given that diameter d = 0.50 mm, we get A = π(0.25 mm)2 = π(0.00025 m)2.

Now, we rearrange Young's modulus formula to solve for ΔL:

E = F / A / (ΔL / L)

ΔL = FL / (EA)

Plugging in the values, with F = 4.0 N, L = 60 mm = 0.060 m, and E = 1×109 N/m2:

ΔL = (4.0 N × 0.060 m) / (1×109 N/m2 × π × (0.00025 m)2)

After calculating, we find the change in length. We can also discuss the reasonableness by comparing this calculated change in length with common experiences using pencil leads.