Final answer:
The energy stored in a wire with a length of 4.0 m and a radius of 1 mm, which is extended by 2.5 mm due to a load of 200.0 N, is calculated using Hooke's Law and the formula for elastic potential energy. The spring constant (k) is determined to be 80,000 N/m, and the energy stored (E) is found to be 0.25 joules.
Step-by-step explanation:
The question requires us to calculate the energy stored in a wire due to elongation. This is a physics problem involving elasticity and mechanical energy. We'll use Hooke's Law and the formula for elastic potential energy. The energy (E) stored in an extended wire is given by the formula:
E = (1/2) * k * x^2
where k is the spring constant of the wire, and x is the extension of the wire. To find the spring constant (k), we use Hooke's Law, which relates force (F) to the extension of the spring (x) as follows:
k = F / x
Here, the force applied is 200.0 N and the extension x is 2.5 mm (or 0.0025 m). First, we calculate the spring constant (k):
k = 200.0 N / 0.0025 m = 80,000 N/m
Next, we use the spring constant to calculate the elastic potential energy:
E = (1/2) * 80,000 N/m * (0.0025 m)^2
E = 0.25 J
Therefore, 0.25 joules of energy is stored in the wire when it is extended by 2.5 mm due to the load.