Question

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

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

Answer 1

Answer: (B)

$2.450$ joule

Since the spring obeys Hooke's law, thus F=kx, or $K=\frac{F}{x}=\frac{4\times 9.8}{2\times {10}^{-2}}=19.6\times {10}^2$
Thus, work done can be written as stored potential energy=$\frac{1}{2}k{x}^2$
=$\frac{1}{2}19.6\times {10}^2\times (0.05{)}^2=2.45J$