Question 268: Properties of Matter

Question

The elastic potential energy stored in a steel wire of length $$20 \mathrm{~m}$$ stretched through $$2 \mathrm{~cm}$$ is $$80 \mathrm{~J}$$. The cross sectional area of the wire is __________ $$\mathrm{mm}^{2}$$.
$$\left(ight.$$ Given, $$\left.y=2.0 	imes 10^{11} \mathrm{Nm}^{-2}ight)$$

Accepted Answer

Given, energy per unit volume = $$\frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain}$$
The stress can be given as $$	ext{stress} = Y 	imes 	ext{strain}$$, where Y is the Young's modulus.
The energy stored in the wire can be written as:
$$	ext{Energy} = \frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain} 	imes 	ext{volume}$$
Substituting the stress formula, we get:
$$	ext{Energy} = \frac{1}{2} 	imes Y 	imes 	ext{strain}^2 	imes A 	imes L$$
We are given that the energy stored is $$80 \ 	ext{J}$$, the original length of the wire is $$20 \ 	ext{m}$$, the elongation is $$2 \ 	ext{cm}$$, and the Young's modulus is $$2.0 	imes 10^{11} \ 	ext{Nm}^{-2}$$. We need to find the cross-sectional area (A) of the wire.
$$80 = \frac{1}{2} 	imes 2 	imes 10^{11} 	imes \left(\frac{2 	imes 10^{-2}}{20}ight)^2 	imes A 	imes 20$$
Now we can solve for A:
$$A = \frac{80 	imes 20^2}{(2.0 	imes 10^{11}) 	imes (2 	imes 10^{-2})^2} = 40 	imes 10^{-6} \ 	ext{m}^2$$
To convert the area to $$	ext{mm}^2$$, we multiply by $$10^6$$:
$$A = 40 	imes 10^{-6} 	imes 10^6 = 40 \ 	ext{mm}^2$$
So, the cross-sectional area of the wire is $$40 \ 	ext{mm}^2$$.

Properties of Matter

The elastic potential energy stored in a steel wire of length $20 \mathrm{~m} stretched through 2 \mathrm{~cm} is 80 \mathrm{~J}. The cross sectional area of the wire is __________ \mathrm{mm}^{2}. \left(\right. Given, \left.y=2.0 \times 10^{11} \mathrm{Nm}^{-2}\right)

Integer / Numeric Answer Type