Question

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 * 1011 Nm-2. Find
1) The energy density of wire.
2) The elastic potential energy stored in the wire.

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Answer 1

Explanation:

Here, l=4.0m;Δl=2×10

−3

m;a=2.0×10

−6

m

2

,Y=2.0×10

11

N/m

2

(i) The energy density of stretched wire

u=

2

1

× stress × strain

=

2

1

×Y×(strain)

2

=

2

1

×2.0×10

11

×(2×10

−3

)/4)

2

=0.25×10

5

=2.5×10

4

J/m

3

.

(ii) Elastic potential energy = energy density × volume

=2.5×10

4

×(2.0×10

−6

)×4.0J=20×10

−2

=0.20J.

Answer 2

Here,

l = 4.0 m ∆l = 2× 10^-3 m,

a = 2.0 × 10^-6

Y = 2.0 × 10^11 Nm^-2

[i]____?

The Energy density of stretched wire

u = U/V = 1/2 stress × strain

= 1/2 × Y × (strain)^2

= 1/2 × 2.0 × 10^11 Nm^-2 [ 2×10^-3/ 4]^2

= 2.5 × 10^4 J/m^3

[ii]____?

Elastic potential energy= energy density×volume

= (2.5 × 10^4 J/m^3) × (8.0 × 10^-6 m^3)

= 0.20 J

__________________________________