Part c if exena applies a force of 220 n at the edge of the door and perpendicular to it, how much time t does it take her to close the door?

The initial formula is J·α = F·w but we need to find the angular acceleration, so the other formula is α = F·w/J

where

J moment of inertia, α angular acceleration, w width of the door ( = length of lever arm of the force F)

Moment of inertia of plane rectangular object, rotating around axis through on edge (see link)

J = (1/3) ·m·w

since mass equals weight W divided by acceleration due to earth gravity g

J = (1/3) · (W/g) ·w²

So the angular acceleration of the door is:

α = F·3·g / (W·w)

= 220N · 3 · 9.81m/s² / (750N · 1.25m)

= 6.90624 s^-2

Integrating twice with initial conditions

ω (t=0) = 0; and

α (t=0) = 0

look for the angular displacement of the door:

α = dω / dt will be ω = ∫ α dt = α·t

ω = dφ / dt will be φ = ∫ ω dt = ∫ α·t dt = (1/2) ·α·t²

So time elapsed until door closed at an angle of

φ' = 90° = (1/2) ·π

is

t' = √ (2·φ'/α)

= √ (π / 6.90624 s^-2)

= 0.67s