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A spring of spring constant $5 \times 10^3 \mathrm{~N} \mathrm{~m}^{-1}$ is stretched initially by $5 \mathrm{~cm}$ from the unstretched position. The work required to stretch it further by another $5 \mathrm{~cm}$ is
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The correct answer is:
$18.75 \mathrm{~N} \mathrm{~m}$
$U_1=\frac{1}{2} k x_1^2=\frac{1}{2} \times\left(5 \times 10^3\right) \times\left(5 \times 10^{-2}\right)^2$
$=6.25 \mathrm{~N} \mathrm{~m}$
$U_2=\frac{1}{2} k x_2^2=\frac{1}{2} \times 5 \times 10^3 \times(5+5)^2 \times 10^{-4}$
$=25 \mathrm{~N} \mathrm{~m}$
$\therefore$ Work done $=U_2-U_1=25.0-6.25$
$=18.75 \mathrm{~N} \mathrm{~m}$
$=6.25 \mathrm{~N} \mathrm{~m}$
$U_2=\frac{1}{2} k x_2^2=\frac{1}{2} \times 5 \times 10^3 \times(5+5)^2 \times 10^{-4}$
$=25 \mathrm{~N} \mathrm{~m}$
$\therefore$ Work done $=U_2-U_1=25.0-6.25$
$=18.75 \mathrm{~N} \mathrm{~m}$
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