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A certain spring that obeys Hooke's law is stretched by an external agent. The work done in stretching the spring by 10 cm is 4 J. How much additional work is required to stretch the spring an additional 10 cm?

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2 J

4 J

8 J

12 J

answer is D.

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Detailed Solution

4.00 J = $\frac{1}{2} k {(0.100 m)}^{2}$ .Therefore, k = 800 N/m and to stretch the spring to 0.200 m requires extra work

$∆ W =$ $\frac{1}{2} (800) {(0.200)}^{2}$ -4.00J = 12.0 J

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