Answer:
6.88 cm
Explanation:
Given that:
mass m of the object = 1.0 kg
Force constant of the spring (k) = 10.1 N/cm = 10.1 × (100) N/m
= 1010 N/m
Radius of the circular path R = 0.515 m
Since 1 revolution = 2 π radian
angular speed of the object ω = 1.84 rev/s = 1.84 × ( 2 π) rad/s
= 11.56 rad/s
Amount of the spring that stretches = X???
However, the amount of the force the spring exerts on the object due to the stretching is directly equal to the centripetal force required by the object for the circular motion:
SO;
[tex]K(X) = m \omega^2 R\\1010*(X) = 1.01 *(11.56)^2 *0.515 \\(X) = \frac{1.01*(11.56)^2*0.515}{1010} \\(X) = 0.0688[/tex]
X = 0.0688 m
Hence, the amount of the spring stretches = 6.88 cm
