This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
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 \, N. The extension produced is e = 0.3 \, mm = 0.3 × 10^-3 \, 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 = (1)/(2) F e Substitute the given values: PE = (1)/(2) × 100 \, N × (0.3 × 10^-3 \, m) PE = 50 × 0.3 × 10^-3 \, J PE = 15 × 10^-3 \, J PE = 0.015 \, J The potential energy in the wire is 0.015 J. b) Step 3: Identify the given values for part (ii). The length of the wire is L' = 2 \, m. The cross-sectional area is A' = 1 \, mm^2 = 1 × (10^-3)^2 \, m^2 = 1 × 10^-6 \, m^2. The extension is e' = 0.1 \, mm = 0.1 × 10^-3 \, m. The Young's Modulus of the wire is Y = 2.0 × 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 = (1)/(2) (A'Y)/(L') (e')^2 Substitute the given values: W = (1)/(2) × (1 × 10^-6 \, m^2) × (2.0 × 10^11 \, Nm^-2)2 \, m × (0.1 × 10^-3 \, m)^2 W = (1)/(2) × 2.0 × 10^5 \, N2 × (1 × 10^-2 × 10^-6 \, m^2) W = (1)/(2) × (1.0 × 10^5 \, N) × (1 × 10^-8 \, m^2) W = 0.5 × 10^-3 \, J W = 0.0005 \, J The work done in stretching the wire is 0.0005 J. Send me the next one 📸