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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

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12.50 N-m

18.75 N-m

25.00 N-m

6.25 N-m

answer is B.

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Detailed Solution

$K = 5 \times 10^{3} N / m W = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = \frac{1}{2} \times 5 \times 10^{3} ({(0.1)}^{2} - {(0.05)}^{2}) = \frac{5000}{2} \times 0.15 \times 0.05 = 18.75 N m$

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