A 700-N girl jumps from a balloon 36.0 m above the ground, and is brought to rest (instantaneously) 4.00 m above the ground by an elastic cord attached to the balloon.
(a) Is the girl’s acceleration upward, downward or zero at this time? Explain.
(b) If the unstretched length of the cord is 25.0 m, what is the cord’s spring constant?
(a) In order to determine the acceleration of the girl, we need to analyze the forces acting on her at the given time. When the girl is brought to rest by the elastic cord, the forces acting on her are her weight (mg) and the force exerted by the elastic cord.
Since the girl is brought to rest, the net force acting on her must be zero. The weight of the girl, mg, acts downward and is given by 700 N. The force exerted by the elastic cord acts in the upward direction. Therefore, the upward force exerted by the cord must be equal in magnitude to the downward force due to her weight.
Since the forces are balanced, the acceleration of the girl at this time is zero. This means that she is neither accelerating upward nor accelerating downward.
(b) To find the cord's spring constant, we can use Hooke's Law which states that the force exerted by an elastic object (such as the cord) is directly proportional to the displacement from its equilibrium position.
Hooke's Law can be represented as F = kx, where F is the force exerted by the cord, k is the spring constant, and x is the displacement of the cord from its unstretched length.
In this case, when the girl jumps from the balloon and reaches the equilibrium point (4.00 m above the ground), the displacement of the cord from its unstretched length is 4.00 m - 25.0 m = -21.0 m. The negative sign indicates that the displacement is in the opposite direction of the positive direction.
Now, we can use the information to solve for the cord's spring constant (k) using Hooke's Law.
F = kx
700 N = k(-21.0 m)
k = -700 N / (-21.0 m)
k ≈ 33.33 N/m (rounded to two decimal places)
Therefore, the cord's spring constant is approximately 33.33 N/m.