Question

A wire of area of cross section 3.0 $m{m}^2$and natural length 50 cm is fixed at one end and a mass of 2.1 kg is hung from the other end. Find the elastic potential energy stored in the wire in steady state. Young's modulus of the material of the wire = $1.9\times {10}^{11}N{m}^{-2}.Takeg=10m{s}^{-2}$

• A. $2\times {10}^{-4}$ J
• B. ${10}^{-4}$ J
• C. $3\times {10}^{-4}$ J
• D. $0.5\times {10}^{-4}$ J

Answer 1

The correct option is A

$2\times {10}^{-4}$ J

The volume of the wire is

V= $3.0m{m}^2$ (50 cm)

$=(3.0\times {10}^{-5}{m}^2)\left(0.50m\right)=1.5\times {10}^{-5}{m}^3$.

Tension in the wire is

T=mg

= (2.1 kg) $10m{s}^{-2}$= 21 N.

The stress = T/A

= $\frac{21N}{3.0m{m}^2}=70\times {10}^{6}N{m}^{-2}$.

The strain = stress/Y

= $\frac{7.0\times {10}^{6}N{m}^{-2}}{1.9\times {10}^{11}N{m}^{-2}}=3.7\times {10}^5$.

The elastic potential energy of the wire is

U =$\frac{1}{2}\left(stress\right)\left(strain\right)\left(volume\right)$

= $\frac{1}{2}(7.0\times {10}^{6}N{n}^{-2})(3.7\times {10}^{-5})(1.5\times {10}^{-5}{m}^3)$

=$1.9\times {10}^{-4}$ J