A steel wire of length 2 m and Young's modulus 2.0 10^11 N m^-2 is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10^-3 respectively, then the elastic potential…

Question

A steel wire of length 2 m and Young's modulus 2.0 10^11 N m^-2 is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10^-3 respectively, then the elastic potential energy density of the wire is 10^5 (in SI units).

Accepted Answer

Correct answer: 25. (25) To find the elastic potential energy density of the steel wire, we need to use the given information and formulae for strain and energy density. Given: The length of the wire, l=2 m Young's modulus, Y=2.0 10^11 N/m^2 Poisson's ratio, =0.2 Transverse strain, rr=10^-3 The formula for Poisson's ratio is: =- ( rr ) ( ll ) From this, we solve for the longitudinal strain ll : ll= 1 ( rr ) Substitute the given values: ll= 10.2 10^-3=5 10^-3 The elastic potential energy density u is given by: u= 12Y _l^2 where _l= ll. Plug in the values: u= 12 2 10^11 (5 10^-3)^2 Simplify further: u= 12 2 10^11 25 10^-6u=25 10^5 ( in SI units ) Thus, the elastic potential energy density of the wire is 25 10^5 SI units.

$A steel wire of length 2 m and Young's modulus 2.0 \times 10^{11} N m^{- 2} is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10^{- 3} respectively, then the elastic potential energy density of the wire is \times 10^{5} (in SI units).$

Solution

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