Question

4. A wire of area of cross-section 3.0 2

and natural length 50 is fixed at one end and

a mass of 2.1 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 × 1011 2 ⁄ .

Take g = 10 2​

Answer 1

Answer:

The volume of the wire is

V=(3.0mm2)(50cm)

=(3.0×10−6m2)(0.50m)=1.5×10−6m3

Tension in the wire is

T=mg

=(2.1kg)(10ms−2)=21N.

The stress =TA

=21N3.0mm2=7.0×106Nm−2

The strain stressY

=7.0×106Nm−21.9×1011Nm,−2=3.7×10−5

The elastic potential energy of the wire is

U=12(stress)(stra∈)(volume)

=12(7.0×106Nm−2)(3.7×10−5)(1.5×10−6m3)

=1.9x10−4J.

Answer 2

Solution :

Volume of the wire

$\tt V = 3.0 \: {mm}^{2} \times 50 \: cm\\ \\ \tt \implies3.0 \times {10}^{ - 6} \: {m}^{2} \times 0.50 \: m \\ \\ \tt \implies1.5 \times {10}^{ - 6} \: {m}^{3}$

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Tension in the wire is

$\large\boxed{\tt T = mg} \\ \\ \tt \implies2.1 \: kg \times 10 \: {m \: s}^{ - 2} \\ \\ \tt \implies 21 \: N$

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$\large \boxed{ \tt The \: stress = \dfrac{ T}{A} }\\ \\ \tt \implies \frac{21 \: n}{3.0 \: {mm}^{2} } \\ \\ \tt \implies 7.0 \times {10}^{6} \: n \: {m}^{ - 2}$

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$\large\boxed{ \tt The \: strain = \dfrac{stress}{Y}} \\ \\ \tt \implies \frac{7.0 \times {10}^{6} \: n \: {m}^{ - 2} }{1.9 \times {10}^{11} \: n \: {m}^{ - 2} } \\ \\ \tt \implies3.7 \times {10}^{6}$

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The elastic potential energy of the wire is

$\boxed{ \tt U = \frac{1}{2} \times stress \times strain \times volume} \\ \\ \tt \implies \frac{1}{2} (7.0 \times {10}^{6})(3.7 \times {10}^{ - 5} )(1.5 \times {10}^{ - 6}) \\ \\ \tt \implies 1.9 \times {10}^{4} \:J \\ \\\large \underline{ \tt \green{Energy= 1.9 \times {10}^{4} \:J}}$