When a 1 kg mass hangs attached to a spring of length 50 cm , the spring stretched 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 the condition, if g= 10 m/s2 .
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We need to get the spring constant of this spring.
Spring constant = k
k = F/e
Where e = The extension in meters
F= Force in netwons.
F = 1 × 10 = 10 N
e = 2/100 = 0.02m
k = 10/0.02 = 500
The spring increases by = x
x = 60 - 50 = 10 cm
In m = 10/100 = 0.1
Elastic energy = 1/2kx²
= 0.5 × 500 × 0.1² = 2.5
= 2.5 Joules
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