Answer:
4.47 m
Step-by-step explanation:
Elastic potential energy is given by 1/2 k x^2
k = spring constant = .5 N/m
and x is the amount of stretch or compression
5 j = 1/2 ( .5 N/m ) x^2
5 / ( 1/2 *.5) = x^2
x = 4.47 m
Spring extension:
Δx ≈ 4.5 m
Given:
k = 0.5 N/m
E = 5 J
__________
Δx - ?
Potential energy of a compressed (stretched) spring:
E = k·(Δx)² / 2
Δx = √ (2·E / k)
Δx = √ (2·5 / 0.5 ) ≈ 4.5 m
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