a wire suspended vertically from one of its ends to stretched by attaching a weight of 200 Newton to the low end . the which stretches the wire by 1 millimetre then the elastic energy stored in the wire is
Answers (2)
Student Expert
3rd Jul, 2020
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!
Comments (0)
3rd Jul, 2020
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!!
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