Mark jumps from a rest at a height in Bungee Jumping.
When he falls 15m from O to A. The elastic string starts to strech.
When he gets to the lowest point B, the elstic string was elongated by 85m.
Then he rebound and move upwards. Give that the mass of Mark is 50kg.
Air resistance can be neglected.
Q1: Find the elastic potential energy in the string when Mark reaches B.
Q2: Is the net force acting on Mark equal to zero when he is at B?
Explain briefly.
Q3: Find the average tension in the elastic string as Mark is moving from A to B
Phy about KE and PE
2011-06-25 10:21 pm
回答 (2)
2011-06-25 10:52 pm
✔ 最佳答案
Q1: Potential energy of Mark, referrenced to B, when he is at O= 50g x (15 + 85) J = 50 000 J
where g is the acceleration due to gravity, taken to be 10 m/s2
Hence, by conservation of energy, elastic potential energy in the string
= 50 000 J
Q2: No, since Mark would be bounced back by the net force, .
the elastic force given the the stretched string is larger than the weight of Mark.
Q3: Using work done = change of energy
hence, F x 85 = 50 000, where F is the average tension in the stringi,e, F = 50000/85 N = 588 N
2011-06-25 10:28 pm
收錄日期: 2021-04-29 17:40:17
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