A daredevil plans to bungee jump from a balloon 52.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 15.0 m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke's law. In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.80 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon.
(a) What length of cord should he use?
(b) What maximum acceleration will he experience?
To solve this problem, we can use the concept of potential energy, Hooke's law, and the conservation of mechanical energy.
(a) To determine the length of cord the daredevil should use, we need to calculate the change in potential energy during the jump.
The potential energy before the jump is given by the formula:
PE_initial = m * g * h_initial
where m is the daredevil's mass, g is the acceleration due to gravity, and h_initial is the initial height (52.0 m).
The potential energy at the stopping point is given by:
PE_final = m * g * h_final
where h_final is the stopping height (15.0 m).
The energy stored in the cord is equal to the difference in potential energy: PE_cord = PE_initial - PE_final.
According to Hooke's law, the potential energy stored in a spring-like system is given by:
PE_cord = (1/2) * k * x^2
where k is the spring constant and x is the displacement from the equilibrium position.
In this case, the displacement x is the difference between the unstretched and stretched lengths of the cord:
x = original length - stretched length
x = L - (5.00 m + 1.80 m)
where L is the length of the cord we need to find.
Since the daredevil is considered a particle, we can neglect his mass in the calculations.
Setting the potential energy equations equal, we have:
(1/2) * k * x^2 = m * g * (h_initial - h_final)
Substituting the values given:
(1/2) * k * (L - (5.00 m + 1.80 m))^2 = 0 * 9.81 m/s^2 * (52.0 m - 15.0 m)
Simplifying:
(1/2) * k * (L - 6.80 m)^2 = 0
Since k and x^2 cannot be zero, we can conclude that L - 6.80 m = 0:
L = 6.80 m
Therefore, the daredevil should use a cord length of 6.80 m.
(b) To find the maximum acceleration experienced by the daredevil during the jump, we can use the principle of conservation of mechanical energy.
The potential energy at the starting point is:
PE_initial = m * g * h_initial
The kinetic energy at the stopping point is zero since he comes to a stop.
The change in potential energy is equal to the change in kinetic energy:
PE_initial - 0 = (1/2) * m * v^2
Simplifying:
m * g * h_initial = (1/2) * m * v^2
The mass of the daredevil cancels out. Solving for v, the velocity at the stopping point:
v = √(2 * g * h_initial)
Substituting the given values:
v = √(2 * 9.81 m/s^2 * (52.0 m - 15.0 m))
Calculating:
v ≈ √(2 * 9.81 m/s^2 * 37.0 m)
v ≈ √(2 * 363.27 m^2/s^2)
v ≈ √(726.54 m^2/s^2)
v ≈ 26.96 m/s
The maximum velocity at the stopping point is approximately 26.96 m/s.
To find the maximum acceleration, we need to consider that the velocity is changing from the moment he starts falling until he comes to a stop. Therefore, acceleration is given by the equation:
a = (v^2 - u^2) / (2 * h)
where u is the initial velocity (0 m/s) and h is the distance traveled during the deceleration (from 52.0 m to 15.0 m).
Substituting the values:
a = (26.96 m/s^2 - 0 m/s^2) / (2 * (52.0 m - 15.0 m))
a = 26.96 m/s^2 / (2 * 37.0 m)
a ≈ 0.366 m/s^2
Therefore, the daredevil will experience a maximum acceleration of approximately 0.366 m/s^2 during the jump.