Question

A mass of 10kg is suspended from the end of a filled rock of length 2m and radius 1mm. What is the elongation of the rock beyond it original length

Answer 1

Explanation:

Answer:

ΔL = 3.12 x 10⁻⁴ m = 0.312 mm

Explanation:

First, we will find the stress applied by the mass on the rod:

\sigma = \frac{F}{A}σ=

A

F

where,

σ = stress = ?

F = Force applied by the mass = weight = mg = (10 kg)(9.81 m/s²) = 98.1 N

A = cross-sectional area = πr² = π(1 mm)² = π(0.001 m)² = 3.14 x 10⁻⁶ m²

Therefore,

\begin{gathered}\sigma = \frac{98.1\ N}{3.14\ x\ 10^{-6}\ m^2}\\\\\end{gathered}

σ=

3.14 x 10

−6

m

2

98.1 N

σ = 3.12 x 10⁷ Pa

Now, we will find the value of strain:

\begin{gathered}E = \frac{\sigma}{\epsilon}\\\\\epsilon = \frac{\sigma}{E}\\\\\epsilon = \frac{3.12\ x\ 10^7\ Pa}{200\ x\ 10^9\ Pa}\end{gathered}

E=

ϵ

σ

ϵ=

E

σ

ϵ=

200 x 10

9

Pa

3.12 x 10

7

Pa

∈ = 1.56 x 10⁻⁴

Now, the change in length can be given by the formula of strain:

\begin{gathered}\epsilon = \frac{\Delta L}{L}\\\\\Delta L = (\epsilon)(L)\end{gathered}

ϵ=

L

ΔL

ΔL=(ϵ)(L)

where,

L = Original Length = 2 m

ΔL = Elongation = ?

Therefore,

ΔL = (1.56 x 10⁻⁴)(2 m)

ΔL = 3.12 x 10⁻⁴ m = 0.312 mm