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!
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)
To find the displacement of the beam, we can use Hooke's Law which states that the work done on a spring is equal to the product of the spring constant and the square of the displacement.
The formula for work done on a spring is given by W = (1/2)kx^2, where W represents the work done, k is the spring constant, and x is the displacement.
Given that the work done is 5 J and the spring constant is 10^6 N/m, we can rearrange the equation to solve for x:
5 = (1/2) * (10^6) * x^2
To solve for x, divide both sides of the equation by (1/2) * (10^6):
5 / [(1/2) * (10^6)] = x^2
Simplifying the equation gives:
10 / (10^6) = x^2
Taking the square root of both sides gives:
x = sqrt(10 / (10^6))
x ≈ 0.00316 m
Therefore, the displacement of the beam is approximately 0.00316 m, which is equivalent to 3.16 mm.
The correct answer is A. 3.2 mm.