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A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is

Option 1)

16 J

Option 2)

8 J

Option 3)

32 J

Option 4)

24 J

Answers (2)

best_answer

As we learnt in

Potential Energy -

U_{f}-U_{i}= \int_{r_{i}}^{r_{f}}\vec{f}\cdot \vec{ds}

- wherein

U_{f}-final\: potential\: energy

U_{i}-initial \: potential\: energy

f-force

ds-small \: displacement

r_{i}-initial \: position

r_{f}-final\: position

 

 Work done = change in energy stored in the spring.

=uf - ui

\omega=\frac{1}{2}kx_{2}^{2}-\frac{1}{2}kx_{i}^{2}

    =\frac{1}{2}\times800\times (15\times 10^{-2})^{2}-\frac{1}{2}\times 800\times(5\times10^{-2})^{2J}

    =400\times10^{-4}(225-25)=8J

Correct answer is 2


Option 1)

16 J

This is an incorrect option.

Option 2)

8 J

This is the correct option.

Option 3)

32 J

This is an incorrect option.

Option 4)

24 J

This is an incorrect option.

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