stored potential energy = (1/2) k X^2
= 5 J
k = 10^6 N/m
X^2 = 10^-5 m^2
X = 3.16*10^-3 m -> 3.2 mm (rounded off)
The work required to bend a steel beam with a spring constant of 10^6 n/m is 5 j. If it is treated like a perfect spring, what is the displacement of the beam?
A. 3.2 mm b. .03 m C. 2.5m D. 0.026 m e. 0
Pls show work thank u!
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