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A spring of spring constant $5 \times 10^3 \mathrm{~N} / \mathrm{m}$ is stretched initially by $5 \mathrm{~cm}$ from the unstretched position. Then the work required to stretch it further by another $5 \mathrm{~cm}$ is
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Verified Answer
The correct answer is:
$18.75 \mathrm{~N}-\mathrm{m}$
$18.75 \mathrm{~N}-\mathrm{m}$
Required work done
$=\frac{1}{2} \mathrm{~K}\left(\mathrm{x}_2^2-\mathrm{x}_1^2\right)=\frac{1}{2} \times 5 \times 10^3\left[10^2-5^2\right] \times 10^{-4}$
$=\frac{1}{2} \times 5 \times 75 \times 10^3 \times 10^{-4}=18.75$
$=\frac{1}{2} \mathrm{~K}\left(\mathrm{x}_2^2-\mathrm{x}_1^2\right)=\frac{1}{2} \times 5 \times 10^3\left[10^2-5^2\right] \times 10^{-4}$
$=\frac{1}{2} \times 5 \times 75 \times 10^3 \times 10^{-4}=18.75$
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