Question

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

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

Answer 1

Answer: (B)

$2.450joule$

$K=\frac{F}{x}=\frac{40}{2\times {10}^{-2}}=19.6\times {10}^{2}N/m$
Work done = $\frac{1}{2}K{x}^2=\frac{1}{2}\times (19.6\ast {10}^2)\times (0.05{)}^2=2.45J$