A steel wire 4.0 m in length is stretched through 2.0 mm. The cross-sectional area of the wire is 2.0 mm2. If Young's modulus of steel is 2.0×1011 N/m2. Find(a) the energy density of wire,(b) the elastic potential energy stored in the wire.

Here, l=4.0 m,Δl=2×10−3 m,A=2.0×10−6 m2,Y=2.0×1011 N/m2(a) The energy density of stretched wireU=21× stress × strain =21×Y×( strain )2=21×2.0×1011×(4(2×10−3))2=0.25×105=2.5×104 J/m3(b) Elastic potential energy = energy density × volume=2.5×104×(2.0×10−6)×4.0 J=20×10−2=0.20 J

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Class 11

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Mechanical Properties of Solids

Youngs modulus and breaking stress

A steel wire 4.0 {~m} in length is stretched through 2.0 {~mm}

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A steel wire $4.0 m$ in length is stretched through $2.0 mm$ . The cross-sectional area of the wire is $2.0 mm^{2}$ . If Young's modulus of steel is $2.0 \times 1 0^{11} N / m^{2}$ . Find
(a) the energy density of wire,
(b) the elastic potential energy stored in the wire.

Views: 6,076 students

Updated on: Mar 22, 2023

Book:

Mechanics Volume 2 (DC Pandey)

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Olivia Discussed

A steel wire

4.0 m

in length is stretched through

2.0 mm

. The cross-sectional area of the wire is

2.0 mm^{2}

. If Young's modulus of steel is

2.0 \times 1 0^{11} N / m^{2}

. Find
(a) the energy density of wire,
(b) the elastic potential energy stored in the wire.

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Here,

l = 4.0 m, Δ l = 2 \times 1 0^{- 3} m, A = 2.0 \times 1 0^{- 6} m^{2}, Y = 2.0 \times 1 0^{11} N / m^{2}

(a) The energy density of stretched wire

U = \frac{1}{2} \times stress \times strain = \frac{1}{2} \times Y \times (strain)^{2} = \frac{1}{2} \times 2.0 \times 1 0^{11} \times (\frac{( 2 \times 1 0 ^{- 3} )}{4})^{2} = 0.25 \times 1 0^{5} = 2.5 \times 1 0^{4} J / m^{3}

(b) Elastic potential energy

=

energy density

\times

volume

= 2.5 \times 1 0^{4} \times (2.0 \times 1 0^{- 6}) \times 4.0 J = 20 \times 1 0^{- 2} = 0.20 J

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Question Text	A steel wire $4.0 m$ in length is stretched through $2.0 mm$ . The cross-sectional area of the wire is $2.0 mm^{2}$ . If Young's modulus of steel is $2.0 \times 1 0^{11} N / m^{2}$ . Find (a) the energy density of wire, (b) the elastic potential energy stored in the wire.
Updated On	Mar 22, 2023
Topic	Mechanical Properties of Solids
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 5
Upvotes	602
Avg. Video Duration	7 min

A steel wire $4.0 m$ in length is stretched through $2.0 mm$ . The cross-sectional area of the wire is $2.0 mm^{2}$ . If Young's modulus of steel is $2.0 \times 1 0^{11} N / m^{2}$ . Find
(a) the energy density of wire,
(b) the elastic potential energy stored in the wire.

Text solutionVerified

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A steel wire 4.0 m in length is stretched through 2.0 mm. The cross-sectional area of the wire is 2.0 mm2. If Young's modulus of steel is 2.0×1011 N/m2. Find(a) the energy density of wire,(b) the elastic potential energy stored in the wire.

Text solutionVerified

Filo tutor solutions (5)

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A steel wire $4.0 m$ in length is stretched through $2.0 mm$ . The cross-sectional area of the wire is $2.0 mm^{2}$ . If Young's modulus of steel is $2.0 \times 1 0^{11} N / m^{2}$ . Find
(a) the energy density of wire,
(b) the elastic potential energy stored in the wire.