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.

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 $k_2 = 28 \text{ N/m}$

When two springs are joined end-to-end (in series), their natural lengths add up.

$$L_{0,total} = L_{0,1} + L_{0,2}$$ $$L_{0,total} = 1.6 \text{ m} + 1.4 \text{ m}$$ $$L_{0,total} = 3.0 \text{ m}$$

The combined spring is stretched between two points 4 m apart.
Total stretched length $L_{stretched} = 4 \text{ m}$
Total extension $x_{total} = L_{stretched} - L_{0,total}$

$$x_{total} = 4 \text{ m} - 3.0 \text{ m}$$ $$x_{total} = 1.0 \text{ m}$$

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 $\boxed{\text{11.67 N}}$.