Question

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 $/$ sexc $\left. ^ { 2 } \right)$

• A. 4.900 joule
• B. 2.450 joule
• C. 0.495 joule
• D. 0.245 joule

Answer 1

Answer: (B)

2.450 joule

Answer 2

$= \frac { 40 } { 2 \times 10 ^ { - 2 } } = 0.2 \mathrm {~N} / \mathrm { m }$

Work done