A 53.0 kg circus performer oscillates up and down at the end of a long elastic rope at a rate of once every 6.60 s. The elastic rope obeys Hooke's Law. By how much is the rope extended beyond its unloaded length when the performer hangs at rest?
omega = sqrt (k/m)
so
2 pi f = sqrt (k/m)
so
2 pi/T = sqrt (k/m)
so
2 pi/ 6.60 = sqrt (k/53)
solve for k
then
m g = k x
x = 53*9.81/k
To determine the extension of the rope when the performer hangs at rest, we can use Hooke's Law, which states that the force exerted by a spring or elastic material is directly proportional to the displacement from its equilibrium position.
Hooke's Law can be expressed as:
F = -kx
Where:
F is the force exerted by the spring or elastic material,
k is the spring constant (a measure of its stiffness),
x is the displacement or extension of the spring from its equilibrium position.
In this case, when the performer hangs at rest, the net force acting on them is zero, since they are not accelerating. Therefore, the force due to gravity pulling the performer downwards is balanced by the upward force exerted by the elastic rope.
The formula to calculate the force due to gravity is:
F_gravity = mg
Where:
m is the mass of the performer (53.0 kg),
g is the acceleration due to gravity (approximately 9.8 m/s²).
Since the force exerted by the elastic rope must balance the force due to gravity, we can write:
F_gravity = F_spring
mg = -kx
Now, we can rearrange the equation to solve for the extension x:
x = -mg/k
To find the spring constant k, we need additional information regarding the oscillation of the performer.