A spring has a diameter of 1cm and the original length of 2m, the spring is pulled by a force of 100N. Determine the change in length of the spring given that young modulus of the spring is 5×10^9 N/M^2.
To determine the change in length of the spring, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to the extension or compression of the spring.
Hooke's Law can be given as:
F = k * ΔL
Where:
F = Force applied to the spring
k = Spring constant (also known as the Young's modulus)
ΔL = Change in length of the spring
In this case, we know the diameter of the spring (1 cm or 0.01 m), the original length (2 m), the force applied (100 N), and the Young's modulus (5×10^9 N/m^2). We need to find the change in length (ΔL).
First, let's calculate the area of the cross-section of the spring using the given diameter:
Area = π * (diameter/2)^2
Area = π * (0.01 m / 2)^2
Area = π * (0.005 m)^2
Area = π * 0.000025 m^2
Area = 7.85×10^-8 m^2 (approximately)
Now, let's calculate the spring constant (k) using the Young's modulus and the cross-sectional area:
k = Young's modulus / Area
k = 5×10^9 N/m^2 / 7.85×10^-8 m^2
k ≈ 6.369×10^16 N/m
Finally, let's rearrange Hooke's Law to solve for ΔL:
ΔL = F / k
ΔL = 100 N / 6.369×10^16 N/m
ΔL ≈ 1.5717×10^-15 m
Therefore, the change in length of the spring is approximately 1.5717×10^-15 meters.