Hey!

As we know,

Elastic energy = (1/2) * F* x

F is force =200N , x is length stretched =1mm = 1*10^-3 m

Therefore, E= (1/2) * 200* 1*10^−3= 0.1 Joule

So, elastici energy / work done in stretching the wire by 1 mm using 200 Newton force  is 0.1 joule that is 1* 10^-1 joule.

Hope it helps!

Hello!!

Hope you are doing great!!!

Given;

Force=200 newton

Extension = 1 mm = 0.001 meter

As we know that

Energy = 1/2*force*extension

Energy = 1/2*200N *0.001 m

Energy = 100 N * 0.001 m

Energy = 0.1 Joule

Therefore Elastic Energy stored in the wire is = 0.1 Joule

Hope it helps !!

Thank you!!