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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 J) in this condition, if g = 10 m/s²

Moderate

Solution

Force constant of a spring
$k = \frac{F}{x} = \frac{mg}{x} = \frac{1 \times 10}{2 \times 10^{- 2}} \Rightarrow k = 500 N / m$
Increment in the length = 60 – 50 = 10 cm
$U = \frac{1}{2} {kx}^{2} = \frac{1}{2} 500 {(10 \times 10^{- 2})}^{2} = 2 .5 J$

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