Question

The massof 0.5 kg is suspended from wire, then lenght of wire increases by 3mm the find out work done​

Answer 1

Hi,

Here is the answer to your question:

(5N * 0.003m)/2

 

that is 7.5*10^-3 joule

 

 

work = 1/2 * F. delta(x)

Answer 2

Explanation:

work done = 1/2YAx^2/L

1/2*Y*A(0.003^2)/L

1/2*9*10^-6*YA/L

hope it helps u

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