Question

A body of mass $ 12 \,kg $ is suspended by a coil spring of natural length $ 50 \,cm $ and spring constant $ 2 \times 10^{3} \, N/m $ . The length of the spring after extension will be

• A. $ 0.00588 \,m $
• B. $ 0.0588 \, m $
• C. $ 0.5588\, m $
• D. None of these

Answer 1

Answer: (C)

From Hooke’s law,
$F=kx \left[\because F=mg\right]$
$\therefore mg=kx, x$ = extension in the spring
or $x=\frac{mg}{k}$
$\Rightarrow x=\frac{12\times9.8}{2\times10^{3}}$
or $x=58.8\times10^{-3}\, m$
$x=0.0588\,m$
Hence, the total length of the spring
$=0.50\,m+0.0588\,m$
$=0.5588\,m$