Question

The jumper has a mass of 50 kg, and the bridge's height above the river is 150 m. The rope has an unstretched length of 11 m and stretches a distance of 4 m before bringing the jumper to a stop. Determine the maximum speed of the jumper and the spring constant of the rope. In your calculation, use g

Answer 1

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