We have that a spring has a spring constant of 256 n/m Must be stretched a distance d to give it an elastic potential energy of 48 j
Where
d=0.6123
From the question we are told
Generally the equation for the Elastic potential energy is mathematically given as
[tex]d=0.6123[/tex]
Where
[tex]48=\frac{1}{2}(256)d^2[/tex]
d=0.6123
Therefore
A spring has a spring constant of 256 n/m Must be stretched a distance d to give it an elastic potential energy of 48 j
Where
d=0.6123
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