Question

When a 1.0 kg mass hangs attached to a spring of length 50 cm, the spring stretches by 2 cm. The mass is pulled down until the length of the spring becomes 60 cm. What is the amount of elastic energy stored in the spring in this condition, if g = $10m/s^2$

• A. 1.5 Joule
• B. 2.0 Joule
• C. 2.5 Joule
• D. 3.0 Joule

Answer 1

The correct option is C

2.5 Joule

Force constant of a spring

$k=\frac{F}{x}=\frac{mg}{x}=\frac{1\times 10}{2\times {10}^{-2}}\Rightarrow k=500N/m$

Increment in the length = 60 - 50 = 10 cm

$U=\frac{1}{2}k{x}^2=\frac{1}{2}500(10\times {10}^{-2}{)}^2=2.5J$