Answer:
h = 24.81 m
Explanation:
Given:-
- The mass of the student, m = 60 kg
- The length of the spring, L = 15 m
- The spring constant, k = 60 N/m
Find:-
How far below the bridge is he hanging
Solution:-
- First realize that after the student attempts a bungee jump he oscillates violently ( dynamic motion ). After some time all the kinetic energy has been converted to Elastic and gravitational potential energy student is (stationary) and hanging down on one end of the spring.
- We will apply equilibrium condition on the student. We see that there are two forces acting on the student. The weight (W) of the student acting downward is in combat with the spring restoring force (Fs) acting upwards.
- Apply equilibrium condition in vertical direction:
Fs - W = 0
Fs = W
- The weight and spring force can be expressed as:
k*x = m*g
Where, g : Gravitational acceleration constant = 9.81 m/s^2
x : The extension of the spring from original position
- Solve for the extension (x) of the spring for this condition.
x = m*g / k
- Plug in the values and evaluate:
x = (60 kg)*(9.81 m/s^2) / (60 N/m)
x = 9.81 m
- The spring extends for about 9.81 m from its original length. So the distance (h) from edge of the bridge would be:
h = L + x
h = 15 + 9.81
h = 24.81 m
