A person pushes on a doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is located 0.800 m from axis of the hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s². What is the moment of inertia of the door about the hinges? A)2.00 kg∙m² B)1.00 kg∙m² C)12.5 kg∙m² D)8.00 kg∙m² E)6.40 kg∙m²

Question

Quizplus · Accepted Answer

The moment of inertia of the door about the hinges is given by:$$I = \frac{F \cdot d}{\alpha}$$where:- F is the force applied to the doorknob (5.00 N)- d is the distance from the doorknob to the axis of the hinges (0.800 m)- α is the angular acceleration of the door (2.00 rad/s^2)Plugging in these values, we get:$$I = \frac{5.00 N \cdot 0.800 m}{2.00 rad/s^2} = 2.00 kg∙m^2$$

A Person Pushes on a Doorknob with a Force of 5.00