Weighty Rope. ?If in Example 15.3 (Section 15.4) we do ?not ?neglect the weight of the rope, what is the wave speed (a) at the bottom of the rope; (b) at the middle of the rope; (c) at the top of the rope?

Solution 20E Step 1: In previous problem, provided, mass of the rope, mrope2.00 kg Mass of the box of rocks, mrock20 kg Length of the rope, L = 80 m Mass per unit length of the rope, m/L = = 2 kg / 80 m = 0.025 kg / m Step 2: a) At the bottom of the rope, the mass of the rope = L L = 80 m at bottom. Therefore, M = 0.025 kg / m × 80 m = 2 kg Tension on the rope, F = Total mass of the system × acceleration due to gravity Total mass of the system = mass of the rope + mass of the rock = 2 kg + 20 kg = 22 kg 2 Tension on the rope, F = (22 kg) × 9.8 m/s = 215.6 N Therefore, the equation for the speed of the wave, v = F Applying these values in the equation, we get, 215.6 N v = 0.025 kg/m v = 8624 m /s 2 Taking square root on both sides, we get, v = 92.87 m/s Speed of the wave at the bottom of the rope is, 92.87 m/s