Physics - bungee jumping
posted by Farah on .
In a head dip bungee jump from abridge over a river, the bungee cord is fastened to the jumpers ankles. The jumper then steps off and falls towards the river until the cord becomes taut. At that point, the cord begins to slow the jumper's descent, until his head just touches the water. The bridge is 22.0m above the river. The un-stretched length of the cord id 12.2m. The jumper is 1.80m tall and has a mass of 60kg. Determine the:
a) The required value of the spring constant for the jump to be successful.
b) The acceleration of the jumper at the bottom of the descent.
The gravitational potential energy of the jumper is mgh, where m is 60kg and h is 22m - 1.8m or 20.2m.
The PE of the bungee at full length is 1/2kx^2, where x is the difference between stretched and unstretched length, i.e. 20.2m - 12.2m or 8m.
Setting these to equal: 1/2kx^2=mgh then k = 2mgh/x^2
k= 2*60*9.8*20.2/8^2= 371N/m