A body of mass 12 kg is suspended by a coil spring of natural length 50 cm and spring constant 2 × 103 N/m . The length of the spring after extension will be

Question

A body of mass 12 kg is suspended by a coil spring of natural length 50 cm and spring constant 2 × 103 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

Tardigrade Admin · Accepted Answer

From Hookeâs law, F=kx [∵ F=mg] ∴ mg=kx, x = extension in the spring or x=(mg/k) ⇒ x=(12×9.8/2×103) or x=58.8×10-3 m x=0.0588 m Hence, the total length of the spring =0.50 m+0.0588 m =0.5588 m

Q. A body of mass $12 k g$ is suspended by a coil spring of natural length $50 c m$ and spring constant $2 \times 1 0^{3} N / m$ . The length of the spring after extension will be

$0.00588 m$

$0.0588 m$

$0.5588 m$

None of these

From Hooke’s law,
$F = k x [∵ F = m g]$
$∴ m g = k x, x$ = extension in the spring
or $x = \frac{m g}{k}$
$\Rightarrow x = \frac{12 \times 9.8}{2 \times 1 0 ^{3}}$
or $x = 58.8 \times 1 0^{- 3} m$
$x = 0.0588 m$
Hence, the total length of the spring
$= 0.50 m + 0.0588 m$
$= 0.5588 m$

Q. A body of mass 12kg is suspended by a coil spring of natural length 50cm and spring constant 2×103N/m . The length of the spring after extension will be

0.00588m

0.0588m

0.5588m