2. Right in the heart of China, the Macau Tower, Macau is the highest com- mercial bungee jump in the world with a maximum length of 233 m dur- ing stretch. When the bungee cord (rope) is un-stretch, its length is 180m. If a man weighing 500 N, dropped from the top of Macau tower, moved vertically down, and reach the maximum length of 233m before the cord compressing up, what then is the (a) force constant of the bungee cord, and (b) what is the stopping power of the bungee cord? (assume that air resistance and mass of the bungee cord are negligible) Given: total length - 233 m, un-stretch length = 180m, Ay= Weight of man= m*g= 500N , g = 9.8 m/s m Required: Force constant = ?, stopping power of Bungee cord = ? Execute: Solve for v before the cord stretches: AGPE -mgh = AKE = 0.5m ve = 2gh = 2*(9.8m/s)*(180m) = vauja Solve for net acceleration due to the cord and earth's gravity; anet+ (1/2-1/)/(2Ay) Initial u is the final velocity before it stretches, while the new final vs = 0 at 33.3m/s2, Fnet - maat - Fnet = -k*Ay, k = -Fnet/Ay = Solve for time elapsed to fully stretch the bungee cord; At = (U-1)/(aet) = Solve for the stopping power of the bungee cord: AW = 0.5k*Ay, Patop =AW/A AW = 0.5k*Ay2 = Potop =AW/At =