Answer:
The maximum speed of the jumper is 54.2m/s
The spring constant of the rope is
K=9.19KN/m
Explanation:
Step one
According to hook's law the applied force F is proportional to the extension e provided the elasticity is maintained
Step two
Given that
mass= 50kg
Height of bridge h=150m
Extention e=4m
Step three
To determine the maximum speed of the jumper
Since he is going with gravity we assume g=9.81m/s²
And we apply the equation of motion
V²=U²+2gh
where u= initial velocity of the jumper =0
h=height of bridge
Step four
Substituting we have
V²=0²+2*9.81*150
V²=2943
V=√2943
V=54.2m/s
Step five
To solve for the spring constant
We have to equate to potential energy of the jumper to the energy stored in the spring
Potential energy of the jumper =mgh
Energy stored in the spring =1/2ke²
Hence mgh=1/2ke²
Making k subject of formula we have
K=2mgh/e²
Substituting our data into the expression we have
K=2*50*9.81*150/4²
K=147150/16
K=9196.9N/m
K=9.19KN/m
