a steel wire of 6m is stretched through 3 x 10-3 m. The cross sectional area of the wire is

2 x 10-6m2. If the young's modulus of the steel is 2 x 1011 N/m2, find the elastic energy density of the wire.

Here,
Y = 2×1011 Nm-2
L = 6 m
x = 3×10-3 m
CSA = 2×10-6 m2

FAxL=Y=>F=YALxAgain from law of force on an elastic material F=kxComparing both eqn. we havek=YALPotential energy stored in the wireU=12kx2=>U=YA2Lx2Energy density = UAL=Y2(xL)2=2×10112×(3×1036)2=25 kJ/m3
 

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