Answer:
The maximum velocity is 1.58 m/s.
Explanation:
A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.
Spring constant, K = 100 N/m
mass, m = 0.1 kg
Amplitude, A = 5 cm = 0.05 m
Let the angular frequency is w.
The maximum velocity is
To solve this exercise it is necessary to apply the concepts related to Centripetal and Perimeter acceleration of a circle.
The perimeter of a circle is defined by
Where,
r= radius
While centripetal acceleration is defined by
Where,
v= velocity
r= radius
PART A)
The distance of a body can be defined based on the speed and the time traveled, that is
x = v*t
For our values the distance is equal to
x = 15*115=1725m
The plane when going to make the turn from east to south makes a quarter of the circumference that is
The same route you take is the distance traveled, that is
PART B)
With the radius is possible calculate he centripetal acceleration,
Therefore the radius of the curva that the plane follows in making the turn is 1098.17m with a centripetal acceleration of