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  • I have a doubt, kindly clarify! Work Energy and Power - JEE Main

A spring of spring constant  5\times 10^{3} N/m is stretched initially by  5 cm from the unstretched position.Then the work required to stretch it further by another  5 cm is

Option 1)

12.50 N­m

 Option 2)

18.75 N­m
Option 3)

25.00 N­m

 Option 4)

6.25 N­m

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

 

 As we discussed in

Potential Energy stored in the spring -

U= \frac{1}{2}\: kx^{2}

- wherein

K=spring \: constant

x= elongation or compression of spring from natural position

 

 Spring constant K=5\times 10^{3} N/m

Total work done in stretching the spring from 5 cm to 10 cm is change in potential energy = U_{f}-U_{i}

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

=\frac{1}{2}\times 5\times 10^{3}[(10\times 10^{-2})^{2}-(5\times 10^{-2})^{2}]

=2500\times 10^{-4} \times75

W = 18.75 N - m

Correct answer is option 2.


Option 1)

12.50 N­m

Option 2)

18.75 N­m

Option 3)

25.00 N­m

Option 4)

6.25 N­m

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perimeter

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