Question

A 25.0 kg door is 0.925 m wide. A customer pushes it perpendicular to its face with a 19.2 N force, and creates an angular acceleration of 1.84 rad/s2. At what distance from the axis was the force applied?

Answer 1

$\tau=I\epsilon\to F\cdot x=\frac{1}{3}ma^2\cdot\epsilon \to$

$x=\frac{1}{3F}ma^2\cdot\epsilon=\frac{1}{3\cdot 19.2}\cdot25\cdot0.925^2\cdot1.84=0.627\ (m)$ . Answer