IIT JEE , AIEEE Forum - Physics , Mechanics

rakesh61

A wire suspende vertically from one of its end is streched by attaching a weight of 200 N to the lower end

The weight stretches the wire by 1 mm the elastic energy stored by wire is

I am gettind ans 0.2 by using formula

mgx = 1/2 kx²

^{But the ans is 0.1}

^{Please tell me why i am wrong ratings sure}

gomurali

elastic potential energy stored in the wire is given by U = 1/2 F e.

1/2 * 200* 10^--3 = 0.1m . The energy stored = 1/2 * stress * strain* volume.

substituting the values for stress, strain and volume we will get above equation for U.

jus_look

hey buddy...!!in dis phenomenan always half of the energy is dissipiated as heat n half is stored as elastic pot energy.....

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