Question

A steel wire of length $4.0\text{m}$ is stretched by $2.0\text{mm}$. The cross-sectional area of the wire is $2.0{\text{mm}}^2$. If Young's modulus of steel is $2.0\times {10}^{11}{\text{N/m}}^2$ , then

• A. the energy density of wire is $2.5\times {10}^4{\text{J/m}}^3$.
• B. the elastic potential energy stored in the wire is $0.20\text{J}$.
• C. the energy density of wire is $2.0\times {10}^4{\text{J/m}}^3$.
• D. the elastic potential energy stored in the wire is $0.30\text{J}$.

Answer 1

Answer: (A), (B)

The correct options are
A the energy density of wire is $2.5\times {10}^4{\text{J/m}}^3$.
B the elastic potential energy stored in the wire is $0.20\text{J}$.

Here, $l=4.0\text{m}$;
$\Delta l=2\times {10}^{-3}\text{m};a=2.0\times {10}^{-6}{\text{m}}^2$
$Y=2.0\times {10}^{11}{\text{N/m}}^2$
The energy density of stretched wire

$u=\frac{1}{2}\times stress\times strain=\frac{1}{2}Y{\left(strain\right)}^2$
$=\frac{1}{2}\times 2.0\times {10}^{11}\times (2\times {10}^{-3}/4{)}^2$
$=0.25\times {10}^5=2.5\times {10}^4{\text{J/m}}^3$

Elastic potential energy = energy density $\times$ volume
$=2.5\times {10}^4\times (2.0\times {10}^{-6})\times 4.0\text{J}$
$=20\times {10}^{-2}=0.20\text{J}$