To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.
PART A) Normal Force.
Here,
Normal reaction of the ring is N and velocity of the ring is v
PART B) Acceleration
Negative symbol indicates deceleration.
<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>
Answer:
The value of the spring constant of this spring is 1000 N/m
Explanation:
Given;
equilibrium length of the spring, L = 10.0 cm
new length of the spring, L₀ = 14 cm
applied force on the spring, F = 40 N
extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm
From Hook's law
Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.
F ∝ e
F = ke
where;
k is the spring constant
k = F / e
k = 40 / 0.04
k = 1000 N/m
Therefore, the value of the spring constant of this spring is 1000 N/m
The frequency of the oscillation in hertz is calculated to be 0.00031 Hz.
The frequency of a wave is defined as the number of cycles completed per second while the period refers to the time taken to complete a cycle. The frequency is the inverse of period.
So;
Period(T) = 54 minutes or 3240 seconds
Frequency (f) = T-1 = 1/T = 1/3240 seconds = 0.00031 Hz
Learn more: brainly.com/question/14588679
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