An aluminum (with Young's modulus 7 x 10^10 Pa) rail is 88.0 m long and has cross-sectional area 0.00050 m^2. The rail is fixed at one end and a 4000 N force is applied at the other end. By how much does the bar stretch?
To determine the amount by which an aluminum rail stretches when a force is applied, we can use Hooke's Law, which states that the amount of stretch or compression of a material is directly proportional to the force applied.
Hooke's Law is expressed as:
F = k * ΔL
Where:
F - Force applied (in Newtons)
k - Spring constant or the modulus of elasticity (Young's modulus)
ΔL - Change in length
Given data:
Young's modulus (k) = 7 x 10^10 Pa (Pascals)
Length of the rail (L) = 88.0 m
Cross-sectional area (A) = 0.00050 m^2
Force applied (F) = 4000 N
First, let's find the stress on the rail:
Stress (σ) = Force / Area
σ = F / A
= 4000 N / 0.00050 m^2
= 8 x 10^6 Pa
Now, we can use Hooke's Law to find the change in length (ΔL):
ΔL = F / k
ΔL = 4000 N / 7 x 10^10 Pa
(Note: The unit of the Young's modulus is Pa)
Calculating the result:
ΔL = 5.7 x 10^-5 m
Therefore, the aluminum rail stretches by approximately 5.7 x 10^-5 meters (or 0.057 mm) when a 4000 N force is applied.