Question

When the load on a wire is slowly increased from 3 to 5 kg-wt, the elongation increases

from 0.61 1.02 mm. The work done during the extension of wire is​

Answer 1

Answer:

0.00818 j

Explanation:

work = ΔFΔL = 20x0.409x10∧-3 = 0.00818 j

Answer 2

The work done during elongation is 0.016 J.

Explanation:

Given that,

Load F= 3 kg-wt

Elongation = 0.61 mm

Load F'= 5 kg -wt

Elongation = 1.02 mm

We need to calculate the work done

Using formula of work done

$W_{0.61}=\dfrac{1}{2}\times stress\times strain\times A\times l$

$W_{0.61}=\dfrac{1}{2}\times\dfrac{F}{A}\times\dfrac{\Delta l}{l}\times A\times l$

$W_{0.61}=\dfrac{1}{2}\times F\times\Delta l$

Put the value into the formula

$W_{0.61}=\dfrac{1}{2}\times 3\times9.8\times0.61\times10^{-3}\ J$

$W_{0.61}=8.967\times10^{-3}\ J$

We need to calculate the work done

Using formula of work done

$W_{1.02}=\dfrac{1}{2}\times F\times\Delta l$

Put the value into the formula

$W_{1.02}=\dfrac{1}{2}\times 5\times9.8\times1.02\times10^{-3}\ J$

$W_{1.02}=24.99\times10^{-3}\ J$

We need to calculate the work done during elongation

Using formula of work done

$W_{elongation}=W_{1.02}-W_{0.61}$

Put the value into the formula

$W_{elongation}=24.99\times10^{-3}-8.967\times10^{-3}$

$W_{elongation}=0.016\ J$

Hence, The work done during elongation is 0.016 J.

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Topic : work done

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