# 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

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.

N

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