12a (i). What is meant by resultant force? (ii). Consider four forces acting at a point as shown in figure 3. Find the magnitude and direction of the resultant force.

Here's the solution to question 13b:

a) Step 1: Identify the given values for part (i). The force applied to the wire is $F = 100 \, \text{N}$. The extension produced is $e = 0.3 \, mm = 0.3 \times 10^{-3} \, \text{m}$.

Step 2: Calculate the potential energy stored in the wire. The potential energy (elastic potential energy) stored in a stretched wire is given by the formula: $PE = \frac{1}{2} F e$ Substitute the given values: $PE = \frac{1}{2} \times 100 \, N \times (0.3 \times 10^{-3} \, m)$ $PE = 50 \times 0.3 \times 10^{-3} \, \text{J}$ $PE = 15 \times 10^{-3} \, \text{J}$ $PE = 0.015 \, \text{J}$ The potential energy in the wire is $\boxed{0.015 J}$.

b) Step 3: Identify the given values for part (ii). The length of the wire is $L' = 2 \, \text{m}$. The cross-sectional area is $A' = 1 \, mm^2 = 1 \times (10^{-3})^2 \, m^2 = 1 \times 10^{-6} \, m^2$. The extension is $e' = 0.1 \, mm = 0.1 \times 10^{-3} \, \text{m}$. The Young's Modulus of the wire is $Y = 2.0 \times 10^{11} \, Nm^{-2}$.

Step 4: Calculate the work done in stretching the wire. The work done in stretching a wire is equal to the elastic potential energy stored, which can be expressed using Young's Modulus: $W = \frac{1}{2} \frac{A'Y}{L'} (e')^2$ Substitute the given values: $W = \frac{1}{2} \times \frac{(1 \times 10^{-6} \, m^2) \times (2.0 \times 10^{11} \, Nm^{-2})}{2 \, m} \times (0.1 \times 10^{-3} \, m)^2$ $W = \frac{1}{2} \times \frac{2.0 \times 10^{5} \, N}{2} \times (1 \times 10^{-2} \times 10^{-6} \, m^2)$ $W = \frac{1}{2} \times (1.0 \times 10^{5} \, N) \times (1 \times 10^{-8} \, m^2)$ $W = 0.5 \times 10^{-3} \, \text{J}$ $W = 0.0005 \, \text{J}$ The work done in stretching the wire is $\boxed{0.0005 J}$.

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