When a 4 kg mass is hung vertically on a light spring that obeys … - Vidyadip

MCQ

When a $4\ kg$ mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by $2\ cms$. The work required to be done by an external agent in stretching this spring by $5\ cms$ will be $\left(g=9.8\right.$ metres $\left./ \operatorname{sexc}^2\right)$

A
$4.900$ joule
✓
$2.450$ joule
C
$0.495$ joule
D
$0.245$ joule

✓

Answer

Correct option: B.

$2.450$ joule

$ K=\frac{F}{x}=\frac{40}{2 \times 10^{-2}}=0.2 \mathrm{~N} / \mathrm{m}$
$ \text { Work done }=\frac{1}{2} K x^2=\frac{1}{2} \times(0.2) \times(0.05)^2=2.5 \mathrm{~J}$

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