Question

A wire is suspended by one end. At the other end a weight mg is attached. The wire extends by l. The ratio of mechanical energy stored in the stretched wire to the work done by mg is:

Answer 1

Answer:

,

Explanation:

answer is given in the attachment

Answer 2

Given that,

Weight = mg

Extends length = l

The ratio of mechanical energy stored in the stretched wire to the work done by weight

We need to calculate the work done by mg

Using formula of work done

$W=\dfrac{1}{2}\times F\times x$

Where, F = force

x =Stretched

Put the value into the formula

$W=\dfrac{1}{2}mgl$

Hence, The work done by mg is $\dfrac{1}{2}mgl$.

Learn more :

Topic : work done

https://brainly.in/question/6666414