Answer:
The force acting on the spring is 6.162 N.
Explanation:
Given that,
Mass of the airplane, m = 0.5 kg
The airplane fly on a 5 m string in a horizontal circle.
It flies level and at constant speed and makes one revolution every 4.00 s. We need to find the force with which the force in the string to keep the plane flying in a circle. It is given by :
[tex]F=mr\omega^2[/tex]
[tex]\omega=0.25\ rev/s\\\\\omega=1.57\ rad/s[/tex]
[tex]F=0.5\times 5\times (1.57)^2\\\\F=6.162\ N[/tex]
So, the force acting on the spring is 6.162 N.