1. Solve the modulus of elasticity for the previous question. 2.Two springs AB and BC are joined together end to end, to form one long spring. The natural lengths of the separate springs are 1.6m and 1.4m, and their modulus of elasticity 20N and 28N respectively. Find the tension in the combined spring if it is stretched between two points 4m apart.
This economics question tests your understanding of economic models and analysis. The step-by-step answer below applies the relevant framework and explains the reasoning.
ScanSolve AI Answer
1. For the previous question, the term "modulus of elasticity" for a spring refers to its spring constant ($k$).
The spring constant was calculated in Step 1 of the previous solution:
$$k = 8 \text{ N/m}$$
The modulus of elasticity (spring constant) for the previous question is $\boxed{\text{8 N/m}}$.
2. For this problem, we will assume that "modulus of elasticity" with units of Newtons (N) refers to the spring constant ($k$) in N/m, as this is standard for spring problems.
Step 1: Identify the given values for each spring.
For spring AB (spring 1):
Natural length $L_{0,1} = 1.6 \text{ m}$
Spring constant $k_1 = 20 \text{ N/m}$
For spring BC (spring 2):
Natural length $L_{0,2} = 1.4 \text{ m}$
Spring constant $k_2 = 28 \text{ N/m}$
Step 2: Calculate the total natural length of the combined spring.
When two springs are joined end-to-end (in series), their natural lengths add up.
$$L_{0,total} = L_{0,1} + L_{0,2}$$
$$L_{0,total} = 1.6 \text{ m} + 1.4 \text{ m}$$
$$L_{0,total} = 3.0 \text{ m}$$
Step 3: Calculate the total extension of the combined spring.
The combined spring is stretched between two points 4 m apart.
Total stretched length $L_{stretched} = 4 \text{ m}$
Total extension $x_{total} = L_{stretched} - L_{0,total}$
$$x_{total} = 4 \text{ m} - 3.0 \text{ m}$$
$$x_{total} = 1.0 \text{ m}$$
Step 4: Calculate the tension in the combined spring.
When springs are in series, the tension ($T$) is the same in each spring, and the total extension is the sum of individual extensions.
From Hooke's Law, $x = T/k$. So, $x_1 = T/k_1$ and $x_2 = T/k_2$.
$$x_{total} = x_1 + x_2$$
$$x_{total} = \frac{T}{k_1} + \frac{T}{k_2}$$
$$x_{total} = T \left( \frac{1}{k_1} + \frac{1}{k_2} \right)$$
Substitute the known values:
$$1.0 \text{ m} = T \left( \frac{1}{20 \text{ N/m}} + \frac{1}{28 \text{ N/m}} \right)$$
$$1.0 = T \left( \frac{28}{20 \times 28} + \frac{20}{20 \times 28} \right)$$
$$1.0 = T \left( \frac{28 + 20}{560} \right)$$
$$1.0 = T \left( \frac{48}{560} \right)$$
Simplify the fraction:
$$1.0 = T \left( \frac{6}{70} \right)$$
$$1.0 = T \left( \frac{3}{35} \right)$$
Solve for $T$:
$$T = \frac{35}{3} \text{ N}$$
$$T \approx 11.67 \text{ N}$$
The tension in the combined spring is $\boxed{\text{11.67 N}}$.
