A bungee jumper leaps from a bridge and undergoes a series ofoscillations. Assum

Question

A bungee jumper leaps from a bridge and undergoes a series ofoscillations. Assume g = 9.78 m/s2 a) If a 60.0 kg jumper uses a bungee cord that has anunstretched length of 33.0 m and she jumps from a height of 50 mabove a river, coming to rest just a few centimeters above thewater surface on the first downward descent, what is the period ofthe oscillations? Assume the bungee cord follows Hooke's law. b) The next jumper in line has a mass of 80.0 kg. Should hejump using the same cord? Explain. A bungee jumper leaps from a bridge and undergoes a series ofoscillations. Assume g = 9.78 m/s2 a) If a 60.0 kg jumper uses a bungee cord that has anunstretched length of 33.0 m and she jumps from a height of 50 mabove a river, coming to rest just a few centimeters above thewater surface on the first downward descent, what is the period ofthe oscillations? Assume the bungee cord follows Hooke's law. b) The next jumper in line has a mass of 80.0 kg. Should hejump using the same cord? Explain. a) If a 60.0 kg jumper uses a bungee cord that has anunstretched length of 33.0 m and she jumps from a height of 50 mabove a river, coming to rest just a few centimeters above thewater surface on the first downward descent, what is the period ofthe oscillations? Assume the bungee cord follows Hooke's law. b) The next jumper in line has a mass of 80.0 kg. Should hejump using the same cord? Explain.

Explanation / Answer

1/2 k (50 - 33)2 = potential energy stored inbungee cord PEB = 144.5 k m g h = 60 * 9.8 * 50 = 29400 J   potential energyof jumper (to be stored in bungee cord) k * 172 = 980 m         (using k * 172 / 2 = 9.8 * 50 m) = (k / m) = (980 / 289) = 1.84 /sec       = (k /m) for simple harmonic motion (restoring force = k x) f = / (2 )    P = 1 / f = 2 / = 3.41 sec An 80 kg jumper cannot use the cord since m g h for thatjumper is greater than that which can be stored in the cord with an available extension of 17 m

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