A.

12.50 N­m

B.

18.75 N­m

C.

25.00 N­m

D.

6.25 N­m

Answers (1)

Here, W = \frac{1}{2}k\left ( x_{2}^{2} -x_{1}^{2} \right )

W = \frac{1}{2}5\times 10^{3}\left ( (\frac{10}{100})^{2} -(\frac{5}{100})^{2} \right )

\Rightarrow W = 18.75 \: J
Hence the work required to stretch the spring further by another 5cm is 18.75 J

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