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Q.

A spring of spring constant 5 x 10³ 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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a

25.00 J

b

12.50 J

c

6.25 J

d

18.75 J

answer is A.

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Detailed Solution

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$Work done to stretch a spring by x is W = \frac{1}{2} {kx}^{2} here W_{1} = \frac{1}{2} x 5 x 10^{3} {(0.05)}^{2} W_{2} = \frac{1}{2} x 5 x 10^{3} {(0.10)}^{2} additional work = \frac{1}{2} x 5 x 10^{3} (0 . 1^{2} - 0 . 05^{2}) \frac{1}{2} x 5 x 10^{3} x 0.15 x 0.05 = 18.75 J$

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