Asked by john
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!
A. 3.2 mm b. .03 m C. 2.5m D. 0.026 m e. 0
Pls show work thank u!
Answers
Answered by
drwls
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)
= 5 J
k = 10^6 N/m
X^2 = 10^-5 m^2
X = 3.16*10^-3 m -> 3.2 mm (rounded off)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.